CTF: FluxFingers4Future - Evil Corp Solution

For this years hack.lu CTF I felt like creating a challenge. Since I work a lot with TLS it was only natural for me to create a TLS challenge. I was informed that TLS challenges are quite uncommon but nevertheless I thought it would be nice to spice the competition up with something "unusual". The challenge mostly requires you to know a lot of details on how the TLS record layer and the key derivation works. The challenge was only solved by one team (0ops from China) during the CTF. Good job!



So let me introduce the challenge first.

The Challenge

You were called by the incident response team of Evil-Corp, the urgently need your help. Somebody broke into the main server of the company, bricked the device and stole all the files! Nothing is left! This should have been impossible. The hacker used some secret backdoor to bypass authentication. Without the knowledge of the secret backdoor other servers are at risk as well! The incident response team has a full packet capture of the incident and performed an emergency cold boot attack on the server to retrieve the contents of the memory (its a really important server, Evil Corp is always ready for such kinds of incidents). However they were unable to retrieve much information from the RAM, what's left is only some parts of the "key_block" of the TLS server. Can you help Evil-Corp to analyze the exploit the attacker used?

(Flag is inside of the attackers' secret message).


TT = Could not recover

key_block:
6B 4F 93 6A TT TT TT TT  TT TT 00 D9 F2 9B 4C B0
2D 88 36 CF B0 CB F1 A6  7B 53 B2 00 B6 D9 DC EF
66 E6 2C 33 5D 89 6A 92  ED D9 7C 07 49 57 AD E1
TT TT TT TT TT TT TT TT  56 C6 D8 3A TT TT TT TT
TT TT TT TT TT TT TT TT  94 TT 0C EB 50 8D 81 C4
E4 40 B6 26 DF E3 40 9A  6C F3 95 84 E6 C5 86 40
49 FD 4E F2 A0 A3 01 06

Download PCAPNG

If you are not interested in the solution and want to try the challenge on your own first, do not read past this point. Spoilers ahead.

The Solution

So lets analyze first what we got. We have something called a "key_block" but we do not have all parts of it. Some of the bytes have been destroyed and are unknown to us. Additionally, we have a PCAP file with some weird messages in them. Lets look at the general structure of the message exchange first.

So looking at the IP address and TCP ports we see that the attacker/client was 127.0.0.1:36674 and was talking with the Server 127.0.0.1:4433. When looking at the individual messages we can see that the message exchange looked something like this:

ENC HS MESSAGE .... ENC HS MESSAGE ->
<- SERVER HELLO, CERTIFICATE, SERVER HELLO DONE
ENC HS MESSAGE .... ENC HS MESSAGE CCS ENC HS MESSAGE, ENC HS MESSAGE ->
<-CCS, ENC HS MESSAGE
ENC HEARTBEAT ->
<- ENC HEARTBEAT
-> ENC APPLICATION DATA
<- INTERNAL ERROR ... INTERNAL ERROR

So this message exchange appears weird. Usually the client is supposed to send a ClientHello in the beginning of the connection, and not encrypted handshake messages. The same is true for the second flight of the client. Usually it transmits its ClientKeyExchange message here, then a ChangeCipherSpec message and finally its Finished message. If we click at the first flight of the client, we can also see some ASCII text fragments in its messages.

Furthermore we can assume that the message sent after the ChangeCipherSpec from the server is actually a TLS Finished message.

Since we cannot read a lot from the messages the client is sending (in Wireshark at least), we can look at the messages the server is sending to get a better hold of what is going on. In the ServerHello message the server selects the parameters for the connection. This reveals that this is indeed a TLS 1.1 connection with TLS_RSA_WITH_AES_256_CBC_SHA , no compression and the Heartbeat Extension negotiated. We can also see that the ServerRandom is: 1023047c60b420bb3321d9d47acb933dbe70399bf6c92da33af01d4fb770e98c (note that it is always 32 bytes long, the UNIX time is part of the ServerRandom).

Looking at the certificate the server sent we can see that the server used a self-signed certificate for Evil.corp.com with an 800-bit RSA modulus:

00ad87f086a4e1acd255d1d77324a05ea7d250f285f3a6de35b9f07c5d083add5166677425b8335328255e7b562f944d55c56ff084f4316fdc9e3f5b009fefd65015a5ca228c94e3fd35c6aba83ea4e20800a34548aa36a5d40e3c7496c65bdbc864e8f161

and the public exponent 65537.


If you pay very close attention to the handshake you can see another weird thing. The size of the exchanged HeartbeatMessages is highly uneven. The client/attacker sent 3500 bytes, the server is supposed to decrypt these messages, and reflect the contents of them. However, the Server sent ~64000 bytes instead. The heartbeat extension became surprisingly well known in 2014, due to the Heartbleed bug in OpenSSL. The bug causes a buffer over-read on the server, causing it to reflect parts of its memory content in return to malicious heartbeat requests. This is a good indicator that this bug might play a role in this challenge.

But what is this key_block thing we got from the incident response team? TLS 1.1 CBC uses 4 symmetric keys in total. Both parties derive these keys from the "master secret" as the key_block. This key_block is then chunked into the individual keys. You can imagine the key_block as some PRF output and both parties knowing which parts of the output to use for which individual key. In TLS 1.1 CBC the key_block is chunked as follows: The first N bytes are the client_write_MAC key, the next N bytes are the server_write_MAC key, the next P bytes are the client_write key and the last P bytes are the server_write key. N is the length of the HMAC key (which is at the time of writing for all cipher suites the length of the HMAC) and P is the length of the key for the block cipher.

In the present handshake AES-256 was negotiated as the block cipher and SHA (SHA-1) was negotiated for the HMAC. This means that N is 20 (SHA-1 is 20 bytes) and P is 32 (AES-256 requires 32 bytes of key material).

Looking at the given key_block we can chunk it into the individual keys:
client_write_MAC = 6B4F936ATTTTTTTTTTTT00D9F29B4CB02D8836CF
server_write_MAC = B0CBF1A67B53B200B6D9DCEF66E62C335D896A92
client_write = EDD97C074957ADE1TTTTTTTTTTTTTTTT56C6D83ATTTTTTTTTTTTTTTTTTTTTTTT
server_write = 94TT0CEB508D81C4E440B626DFE3409A6CF39584E6C5864049FD4EF2A0A30106

Since not all parts of the key_block are present, we can see that we actually have 14/20 bytes of the client_write_MAC key, the whole server_write_MAC key, 12/32 bytes of the client_write key and 31/32 bytes of the server_write key.

The client_write_MAC key is used in the HMAC computations from the client to the server (the server uses the same key to verify the HMAC),
The server_write_MAC key is used in the HMAC computations from the server to the client (the client uses the same key to verify the HMAC),
The client_write key is used to encrypt messages from the client to the server, while the server_write key is used to encrypt messages from the server to the client.

So looking at the keys we could compute HMAC's from the client if we could guess the remaining 6 bytes. We could compute HMAC's from the server directly, we have not enough key material to decrypt the client messages, but we could decrypt server messages if we brute-forced one byte of the server_write key. But how would you brute force this byte? When do we know when we got the correct key? Lets look at how the TLS record layer works to find out :)

The Record Layer

TLS consists out of multiple protocols (Handshake, Alert, CCS, Application (and Heartbeat)). If one of those protocols wants to send any data, it has to pass this data to the record layer. The record layer will chunk this data, compress it if necessary, encrypt it and attach a "record header" to it.

This means, that if we want to decrypt a message we know that if we used the correct key the message should always have a correct padding. If we are unsure we could even check the HMAC with the server_write_MAC key.

In TLS 1.0 - TLS 1.2 the padding looks like this:

1 byte padding  : 00
2 bytes padding: 01 01
3 bytes padding: 02 02 02
4 bytes padding: 03 03 03 03
...

So if we guessed the correct key we know that the plaintext has to have valid padding.
An ideal candidate for our brute force attack is the server Finished message. So lets use that to check our key guesses.
The ciphertext looks like this:
0325f41d3ebaf8986da712c82bcd4d55c3bb45c1bc2eacd79e2ea13041fc66990e217bdfe4f6c25023381bab9ddc8749535973bd4cacc7a4140a14d78cc9bddd

The first 16 bytes of the ciphertext are the IV:
IV: 0325f41d3ebaf8986da712c82bcd4d55
Therefore the actual ciphertext is:
Ciphertext: c3bb45c1bc2eacd79e2ea13041fc66990e217bdfe4f6c25023381bab9ddc8749535973bd4cacc7a4140a14d78cc9bddd

The 256 key candidates are quick to check, and it is revealed that 0xDC was the missing byte.
(The plaintext of the Finished is 1400000C455379AAA141E1B9410B413320C435DEC948BFA451C64E4F30FE5F6928B816CA0B0B0B0B0B0B0B0B0B0B0B0B)

Now that we have the full server_write key we can use it to decrypt the heartbeat records.

This is done in the same way as with the Finished. Looking at the decrypted heartbeat messages we can see a lot of structured data, which is an indicator that we are actually dealing
with the Heartbleed bug. If we convert the content of the heartbeat messages to ASCII we can actually see that the private key of the server is PEM encoded in the first heartbeat message.

Note: This is different to a real heartbeat exploit. Here you don't usually get the private key nicely encoded but have to extract it using the coppersmith's attack or similar things. I did not want to make this challenge even harder so I was so nice to write it to the memory for you :)


The private key within the Heartbeat messages looks like this:
-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----

We should store it in a file and decode it with OpenSSL to get the actual key material.

>> openssl rsa -in key.pem -text -noout
RSA Private-Key: (800 bit, 2 primes)
modulus:
    00:ad:87:f0:86:a4:e1:ac:d2:55:d1:d7:73:24:a0:
    5e:a7:d2:50:f2:85:f3:a6:de:35:b9:f0:7c:5d:08:
    3a:dd:51:66:67:74:25:b8:33:53:28:25:5e:7b:56:
    2f:94:4d:55:c5:6f:f0:84:f4:31:6f:dc:9e:3f:5b:
    00:9f:ef:d6:50:15:a5:ca:22:8c:94:e3:fd:35:c6:
    ab:a8:3e:a4:e2:08:00:a3:45:48:aa:36:a5:d4:0e:
    3c:74:96:c6:5b:db:c8:64:e8:f1:61
publicExponent: 65537 (0x10001)
privateExponent:
    26:3f:79:3f:64:26:2d:be:6a:96:06:e3:e5:25:c7:
    d7:3b:9f:05:e5:8a:6f:b4:38:a9:54:1d:45:30:24:
    31:55:d8:b9:62:bb:51:9f:56:6b:d9:d8:ba:5c:a3:
    be:0f:51:a1:63:8e:f9:26:83:36:a8:62:4d:62:a8:
    b1:0f:99:ae:f1:b7:03:54:c3:31:c5:59:4a:43:f1:
    6d:70:ae:86:14:69:a7:88:f4:4f:d9:5e:bd:08:b4:
    b3:cb:51:d5:8b:85:cd:51:62:f1
prime1:
    00:d3:4b:ae:02:b6:60:e2:62:0e:5c:ae:ef:cb:7f:
    79:56:c1:7b:2e:26:ba:1d:b1:6b:1b:c4:77:cc:46:
    5f:86:c8:06:54:ee:5a:52:88:ab:03:cd:c4:4f:1f:
    ca:09:af:e2:e1:c5
prime2:
    00:d2:3e:d4:93:54:e1:63:e5:02:61:d7:82:9b:8f:
    cf:79:5a:a4:3d:8b:bb:8c:d4:c7:1b:64:c6:af:f0:
    7a:86:cb:71:c8:5a:57:06:5a:ba:30:94:3f:a8:f7:
    51:5c:b9:0b:16:ed
exponent1:
    00:9f:fc:e2:c2:4d:03:e9:06:24:27:cb:91:e8:3d:
    1a:4c:35:6e:26:d0:ce:05:e3:ab:dd:37:93:1a:0a:
    83:14:53:ea:6f:6e:96:d7:7d:82:37:fc:1a:d3:6a:
    97:99:64:23:5f:9d
exponent2:
    00:9e:d7:cd:6f:4a:8f:c7:13:3c:8b:83:71:2f:ea:
    a5:0b:c0:89:99:de:3a:62:9a:57:9b:c0:b5:c4:33:
    61:be:f9:72:0b:b7:05:4c:cd:bb:21:fc:bf:63:ff:
    06:bf:91:26:69:b9
coefficient:
    00:8e:9f:4d:fb:17:44:9b:87:56:e7:59:72:52:9e:
    9b:a1:55:3a:7f:16:62:99:f6:11:7d:80:0d:53:66:
    d5:16:66:ec:48:05:3d:8f:38:d6:6f:40:6e:5d:ec:
    4a:29:c5:fb:5c:6a

So now we got the private key. But what do we do with it? Since this is an RSA handshake we should be able to decrypt the whole session (RSA is not perfect forward secure). Loading it into Wireshark does not work, as Wireshark is unable to read the messages sent by the client. What is going on there?

De-fragmentation

So if you do not yet have a good idea of what the record layer is for, you can imagine it like envelops. If someone wants to send some bytes, you have to put them in an envelop and transmit them. Usually implementations use one big envelop for every message, however you can also send a single message in multiple envelops.

The attacker did exactly that. He fragmented its messages into multiple records. This is not very common for handshake messages but fine according to the specification and accepted by almost all implementations. However, Wireshark is unable to decode these kinds of messages and therefore unable to use our private key to decrypt the connection. So we have to do this step manually.

So each record has the following fields:
Type | Version | Length | Data
If we want to reconstruct the ClientHello message we have to get all the data fields of the records of the first flight and decode them.
This is simply done by clicking on each record in Wireshark and concatenating the data fields. This step is at least on my Wireshark version (3.0.5) not very easy as the copying is actually buggy, and Wireshark is not copying the correct bytes.

 As you can see in the image, the record is supposed to have a length of 8 bytes, but Wireshark is only copying 4 bytes. I am not sure if this bug is actually only in my version or affects all Wireshark versions. Instead of copying the records individually I therefore copied the whole TCP payload and chunked it manually into the individual records.

16030200080100009e03020000
160302000800000000004e6f62
16030200086f64796b6e6f7773
1603020008696d616361740000
16030200080000000000002053
1603020008746f70206c6f6f6b
1603020008696e67206e6f7468
1603020008696e6720746f2066
1603020008696e646865726500
16030200080200350100005300
16030200080f00010113370015
16030200084576696c436f7270
1603020008206b696c6c732070
1603020008656f706c65000d00
16030200082c002a0102020203
16030200080204020502060201
16030200080102010301040105
16030200080106010103020303
160302000803040305030603ed
1603020008edeeeeefefff0100
16030200020100

If we structure this data it looks like this:
Type  Version Length  Payload
16    0302    0008    0100009e03020000
16    0302    0008    00000000004e6f62
16    0302    0008    6f64796b6e6f7773
16    0302    0008    696d616361740000
16    0302    0008    0000000000002053
16    0302    0008    746f70206c6f6f6b
16    0302    0008    696e67206e6f7468
16    0302    0008    696e6720746f2066
16    0302    0008    696e646865726500
16    0302    0008    0200350100005300
16    0302    0008    0f00010113370015
16    0302    0008    4576696c436f7270
16    0302    0008    206b696c6c732070
16    0302    0008    656f706c65000d00
16    0302    0008    2c002a0102020203
16    0302    0008    0204020502060201
16    0302    0008    0102010301040105
16    0302    0008    0106010103020303
16    0302    0008    03040305030603ed
16    0302    0008    edeeeeefefff0100
16    0302    0002    0100

The actual message is the concatenation of the record payloads:

0100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

So what is left is to parse this message. There is an easy way on how to do this an a labor intensive manual way. Lets do the manual process first :) .
We know from the record header that his message is in fact a handshake message (0x16).
According to the specification handshake messages look like this:
    
      struct {
          HandshakeType msg_type;    /* handshake type */
          uint24 length;             /* bytes in message */
          select (HandshakeType) {
              case hello_request:       HelloRequest;
              case client_hello:        ClientHello;
              case server_hello:        ServerHello;
              case certificate:         Certificate;
              case server_key_exchange: ServerKeyExchange;
              case certificate_request: CertificateRequest;
              case server_hello_done:   ServerHelloDone;
              case certificate_verify:  CertificateVerify;
              case client_key_exchange: ClientKeyExchange;
              case finished:            Finished;
          } body;
      } Handshake;
    
This is RFC speak for: Each handshake message starts with a type field which says which handshake message this is, followed by a 3 byte length field which determines the length of rest of the handshake message.
So in our case the msg_type is 0x01 , followed by a 3 Byte length field (0x00009e, 158[base10]). 0x01 means ClientHello (https://www.iana.org/assignments/tls-parameters/tls-parameters.xhtml#tls-parameters-7). This means we have to parse the bytes after the length field as a ClientHello.
    
      {
          ProtocolVersion client_version;
          Random random;
          SessionID session_id;
          CipherSuite cipher_suites<2..2^16-2>;
          CompressionMethod compression_methods<1..2^8-1>;
          select (extensions_present) {
              case false:
                  struct {};
              case true:
                  Extension extensions<0..2^16-1>;
          };
      } ClientHello;

This means: The next 2 bytes are the ProtocolVersion, the next 32 bytes are the ClientRandom, the next byte is the SessionID Length, the next SessionID Length many bytes are the SessionID, the next 2 bytes are the CipherSuite Length bytes, followed by CipherSuite Length many CipherSuites, followed by a 1 byte Compression Length field, followed by Compression Length many CompressionBytes followed by a 2 byte Extension Length field followed by extension length many ExtensionBytes. So lets try to parse this:

Handshakye Type   : 01
Handshake Length  : 00009e
ProtocolVersion   : 0302
ClientRandom      : 000000000000004e6f626f64796b6e6f7773696d616361740000000000000000
SessionID Length  : 20
SessionID         : 53746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e6468657265
CipherSuite Length: 0002
CipherSuites      : 0035
Compression Length: 01
CompressionBytes  : 00
Extension Length  : 0053
ExtensionBytes:   : 000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

This is manual parsing is the slow method of dealing with this. Instead of looking at the specification to parse this message we could also compare the message structure to another ClientHello. This eases this process a lot. What we could also do is record the transmission of this message as a de-fragmented message to something and let Wireshark decode it for us. To send the de-fragmented message we need to create a new record header ourselves. The record should look like this:

Type   : 16
Version: 0302
Length : 00A2
Payload: 0100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

To send this record we can simply use netcat:


echo '16030200A20100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100' | xxd -r -p | nc localhost 4433


Now we can use Wireshark to parse this message. As we can see now, the weired ASCII fragments we could see in the previous version are actually the ClientRandom, the SessionID, and a custom extension from the attacker. Now that we have de-fragmented the message, we know the ClientRandom: 000000000000004e6f626f64796b6e6f7773696d616361740000000000000000

De-fragmenting the ClientKeyExchange message

Now that we have de-fragmented the first flight from the attacker, we can de-fragment the second flight from the client. We can do this in the same fashion as we de-fragmented the ClientHello.

16    0302    0008    1000006600645de1
16    0302    0008    66a6d3669bf21936
16    0302    0008    5ef3d35410c50283
16    0302    0008    c4dd038a1b6fedf5
16    0302    0008    26d5b193453d796f
16    0302    0008    6e63c144bbda6276
16    0302    0008    3740468e21891641
16    0302    0008    0671318e83da3c2a
16    0302    0008    de5f6da6482b09fc
16    0302    0008    a5c823eb4d9933fe
16    0302    0008    ae17d165a6db0e94
16    0302    0008    bb09574fc1f7b8ed
16    0302    0008    cfbcf9e9696b6173
16    0302    0002    f4b6

14    0302    0001    01

16    0302    0030    cbe6bf1ae7f2bc40a49709a06c0e3149a65b8cd93c2525b5bfa8f696e29880d3447aef3dc9a996ca2aff8be99b1a4157
16    0302    0030    9bf02969ca42d203e566bcc696de08fa80e0bfdf44b1b315aed17fe867aed6d0d600c73de59c14beb74b0328eacadcf9

Note that his time we have 3 record groups. First there is chain of handshake records, followed by a ChangeCipherSpec record, followed by 2 more handshake records. The TLS specification forbids that records of different types are interleaved. This means that the first few records a probably forming a group of messages. The ChangeCipherSpec record is telling the server that subsequent messages are encrypted. This seems to be true, since the following records do not appear to be plaintext handshake messages.

So lets de-fragment the first group of records by concatenating their payloads:

1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b6

Since this is a handshake message, we know that the first byte should tell us which handshake message this is. 0x10 means this is a ClientKeyExchange message. Since we already know that TLS_RSA_WITH_AES_256_CBC_SHA was negotiated for this connection, we know that this is an RSA ClientKeyExchange message.

These messages are supposed to look like this (I will spare you the lengthy RFC definition):

Type (0x10)
Length (Length of the content) (3 bytes)
EncryptedPMS Length(Length of the encrypted PMS) (2 bytes)
EncrpytedPMS  (EncryptedPMS Length many bytes)
    
For our message this should look like this:

Type: 10
Length: 000066
Encrypted PMS Length: 0064
Encrypted PMS: 5de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b6

Now that we got the Encrypted PMS we can decrypt it with the private key. Since the connection negotiated RSA as the key exchange algorithm this is done with:

encPMS^privKey mod modulus = plainPMS

We can solve this equation with the private key from the leaked PEM file.

2445298227328938658090475430796587247849533931395726514458166123599560640691186073871766111778498132903314547451268864032761115999716779282639547079095457185023600638251088359459150271827705392301109265654638212139757207501494756926838535350 ^ 996241568615939319506903357646514527421543094912647981212056826138382708603915022492738949955085789668243947380114192578398909946764789724993340852568712934975428447805093957315585432465710754275221903967417599121549904545874609387437384433 mod 4519950410687629988405948449295924027942240900746985859791939647949545695657651701565014369207785212702506305257912346076285925743737740481250261638791708483082426162177210620191963762755634733255455674225810981506935192783436252402225312097

Solving this equation gives us:

204742908894949049937193473353729060739551644014729690547520028508481967333831905155391804994508783506773012994170720979080285416764402813364718099379387561201299457228993584122400808905739026823578773289385773545222916543755807247900961

in hexadecimal this is:

00020325f41d3ebaf8986da712c82bcd4d554bf0b54023c29b624de9ef9c2f931efc580f9afb081b12e107b1e805f2b4f5f0f1000302476574204861636b6564204e6f6f622c20796f752077696c6c206e65766572206361746368206d65212121212121

The PMS is PKCS#1.5 encoded. This means that it is supposed to start with 0x0002 followed by a padding which contains no 0x00 bytes, followed by a separator 0x00 byte followed by a payload. In TLS, the payload has to be exactly 48 bytes long and has to start with the highest proposed protocol version of the client. We can see that this is indeed the case for our decrypted payload. The whole decrypted payload is the PMS for the connection.

This results in the PMS: 0302476574204861636b6564204e6f6f622c20796f752077696c6c206e65766572206361746368206d65212121212121 (which besides the protocol version is also ASCII :) )

Now that we have the PMS its time to revisit the key scheduling in TLS. We already briefly touched it but here is a overview:

As you can see, we first have to compute the master secret. With the master secret we can reconstruct the key_block. If we have computed the key_block, we can finally get the client_write key and decrypt the message from the attacker.


master secret = PRF ( PMS, "master secret", ClientRandom | ServerRandom)

key_block = PRF (master_secret, "key expansion", ServerRandom |  ClientRandom )

Where "master secret" and "key expansion" are literally ASCII Strings.

Note that in the key_block computation ClientRandom and ServerRandom are exchanged.



To do this computation we can either implement the PRF ourselfs, or easier, steal it from somewhere. The PRF in TLS 1.1 is the same as in TLS 1.0. Good places to steal from are for example openssl (C/C++), the scapy project (python), the TLS-Attacker project (java) or your favourite TLS library. The master secret is exactly 48 bytes long. The length of the key_block varies depending on the selected cipher suite and protocol version. In our case we need 2 * 20 bytes (for the 2 HMAC keys) + 2 * 32 bytes (for the 2 AES keys) = 104 bytes.

I will use the TLS-Attacker framework for this computation. The code will look like this:


This results in the following master secret: 292EABADCF7EFFC495825AED17EE7EA575E02DF0BAB7213EC1B246BE23B2E0912DA2B99C752A1F8BD3D833E8331D649F  And the following key_block:
6B4F936ADE9B4010393B00D9F29B4CB02D8836CFB0CBF1A67B53B200B6D9DCEF66E62C335D896A92EDD97C074957ADE136D6BAE74AE8193D56C6D83ACDE6A3B365679C5604312A1994DC0CEB508D81C4E440B626DFE3409A6CF39584E6C5864049FD4EF2A0A30106

Now we can chunk our resulting key_block into its individual parts. This is done analogously to the beginning of the challenge.

client_write_mac key = 6B4F936ADE9B4010393B00D9F29B4CB02D8836CF
server_write_mac key = B0CBF1A67B53B200B6D9DCEF66E62C335D896A92
client_write key = EDD97C074957ADE136D6BAE74AE8193D56C6D83ACDE6A3B365679C5604312A19
server_write key = 94DC0CEB508D81C4E440B626DFE3409A6CF39584E6C5864049FD4EF2A0A30106

Now that we have the full client_write key we can use that key to decrypt the application data messages. But these messages are also fragmented. But since the messages are now encrypted, we cannot simply concatenate the payloads of the records, but we have to decrypt them individually and only concatenate the resulting plaintext.

Analogue to the decryption of the heartbeat message, the first 16 bytes of each encrypted record payload are used as an IV

IV, Ciphertext Plaintext
6297cb6d9afba63ec4c0dd7ac0184570    a9c605307eb5f8ccbe8bbc210ff1ff14943873906fad3eca017f49af8feaec87      557365723A20726FB181CF546350A88ACBE8D0248D6FF07675D1514E03030303
063c60d43e08c4315f261f8a4f06169a    cdb5818d80075143afe83c79b570ab0b349b2e8748f8b767c54c0133331fb886    6F743B0A50617373D6F734D45FB99850CCAF32DF113914FC412C523603030303
cd839b95954fcadf1e60ee983cbe5c21    ac6f6e1fe34ae4b1214cded895db4746b8e38d7960d7d45cb001aab8e18c7fc7    3A20726F6F743B0A937048A265327642BD5626E00E4BC79618F9A95C03030303
8092d75f72b16cb23a856b00c4c39898    8df099441e10dca5e850398e616e4597170796b7202e2a8726862cd760ebacdf    6563686F20224F7769EACFBEEB5EE5D1F0B72306F8C78AD86CB4835003030303
8e9f83b015fce7f9c925b8b64abee426    224a5fbd2d9b8fc6ded34222a943ec0e8e973bcf125b81f918e391a22b4b0e65    6E6564206279204061736E93BFDC5103C8C2FE8C543A72B924212E8403030303
0e24ba11e41bfcf66452dc80221288ce    a66fb3aed9bdc7e08a31a0e7f14e11ce0983ec3d20dd47c179425243b14b08c9    6963306E7A31223B84A3CAFA7980B461DE0A6410D6251551AE401DD903030303
0465fdb05b121cdc08fa01cdacb2c8f4    eff59402f4dbf35a85cc91a6d1264a895cd1b3d2014c91fbba03ec4c85d058c9     0A7375646F20726DB97422D8B30C54CC672FFEC3E9D771D4743D96B903030303
e2ddbbb83fe8318c41c26d57a5813fab    89549a874ff74d83e182de34ecf55fff1a57008afd3a29ef0d839b991143cd2a      202F202D72663B0A996F3F1789CB9B671223E73C66A0BA578D0C0F3203030303
524f5210190f73c984bd6a59b9cf424c    b7f30fafe5ea3ac51b6757c51911e86b0aa1a6bbf4861c961f8463154acea315    0A666C61677B436868BF764B01D2CDCB2C06EA0DFC5443DABB6EC9AE03030303
32765985e2e594cddca3d0f45bd21f49    a5edfe89fdb3782e2af978585c0e27ba3ef90eb658304716237297f97e4e72bc    696D696368616E67FBF32127FA3AF2F97770DE5B9C6D376A254EF51E03030303
e0ae69b1fa54785dc971221fd92215fb    14e918a9e6e37139153be8cb9c16d2a787385746f9a80d0596580ba22eaf254e    61467233346B7D8BE8B903A167C44945E7676BF99D888A4B86FA8E0404040404

The plaintext then has to be de-padded and de-MACed.

Data HMAC Pad:
557365723A20726F    B181CF546350A88ACBE8D0248D6FF07675D1514E    03030303
6F743B0A50617373    D6F734D45FB99850CCAF32DF113914FC412C5236    03030303
3A20726F6F743B0A    937048A265327642BD5626E00E4BC79618F9A95C    03030303
6563686F20224F77    69EACFBEEB5EE5D1F0B72306F8C78AD86CB48350    03030303
6E65642062792040    61736E93BFDC5103C8C2FE8C543A72B924212E84    03030303
6963306E7A31223B    84A3CAFA7980B461DE0A6410D6251551AE401DD9    03030303
0A7375646F20726D    B97422D8B30C54CC672FFEC3E9D771D4743D96B9    03030303
202F202D72663B0A    996F3F1789CB9B671223E73C66A0BA578D0C0F32    03030303
0A666C61677B4368    68BF764B01D2CDCB2C06EA0DFC5443DABB6EC9AE    03030303
696D696368616E67    FBF32127FA3AF2F97770DE5B9C6D376A254EF51E    03030303
61467233346B7D      8BE8B903A167C44945E7676BF99D888A4B86FA8E    0404040404

This then results in the following data:

Data:
557365723A20726F6F743B0A506173733A20726F6F743B0A6563686F20224F776E656420627920406963306E7A31223B0A7375646F20726D202F202D72663B0A0A666C61677B4368696D696368616E6761467233346B7D8B

Which is ASCII for:

User: root;
Pass: root;
echo "Owned by @ic0nz1";
sudo rm / -rf;

flag{ChimichangaFr34k}


Honestly this was quite a journey. But this presented solution is the tedious manual way. There is also a shortcut with which you can skip most of the manual cryptographic operations.

The Shortcut

After you de-fragmented the messages you can patch the PCAP file and then use Wireshark to decrypt the whole session. This way you can get the flag without performing any cryptographic operation after you got the private key. Alternatively you can replay the communication and record it with Wireshark. I will show you the replay of the messages. To recap the de-fragmented messages looks like this:

ClientHello
0100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

ClientKeyExchange:
1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b6

We should now add new (not fragmented) record header to the previously fragmented message. The messages sent from the server can stay as they are. The ApplicationData from the client can also stay the same. The messages should now look like this

ClientHello
16030200A20100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

ServerHello / Certificate / ServerHelloDone
160302006A1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b61403020001011603020030cbe6bf1ae7f2bc40a49709a06c0e3149a65b8cd93c2525b5bfa8f696e29880d3447aef3dc9a996ca2aff8be99b1a415716030200309bf02969ca42d203e566bcc696de08fa80e0bfdf44b1b315aed17fe867aed6d0d600c73de59c14beb74b0328eacadcf9

ClientKeyExchange / ChangeCipherSpec / Finished
160302006A1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b61403020001011603020030cbe6bf1ae7f2bc40a49709a06c0e3149a65b8cd93c2525b5bfa8f696e29880d3447aef3dc9a996ca2aff8be99b1a415716030200309bf02969ca42d203e566bcc696de08fa80e0bfdf44b1b315aed17fe867aed6d0d600c73de59c14beb74b0328eacadcf9')

ApplicationData
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

What we want to do now is create the following conversation:
CH->
<-SH, CERT, SHD
-> CKE, CCS, FIN
-> APP, APP ,APP

This will be enough for Wireshark to decrypt the traffic. However, since we removed some messages (the whole HeartbeatMessages) our HMAC's will be invalid.

We need to record an interleaved transmission of these message with Wireshark. I will use these simple python programs to create the traffic:




If we record these transmissions and tick the flag in Wireshark to ignore invalid HMAC's we can see the plaintext (if we added the private key in Wireshark).

Challenge Creation

I used our TLS-Attacker project to create this challenge. With TLS-Attacker you can send arbitrary TLS messages with arbitrary content in an arbitrary order, save them in XML and replay them. The communication between the peers are therefore only two XML files which are loaded into TLS-Attacker talking to each other. I then copied parts of the key_block from the debug output and the challenge was completed :)

If you have question in regards to the challenge you can DM me at @ic0nz1
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In this post, we show a proof-of-concept attack that gives us root access to a victim's VM in the Cloud Management Platform OpenNebula, which means that we can read and write all its data, install software, etc. The interesting thing about the attack is, that it allows an attacker to bridge the gap between the cloud's high-level web interface and the low-level shell-access to a virtual machine.

Like the latest blogpost of this series, this is a post about an old CSRF- and XSS-vulnerability that dates back to 2014. However, the interesting part is not the vulnerability itself but rather the exploit that we were able to develop for it.

An attacker needs the following information for a successful attack.

• ID of the VM to attack
  OpenNebula's VM ID is a simple global integer that is increased whenever a VM is instantiated. The attacker may simply guess the ID. Once the attacker can execute JavaScript code in the scope of Sunstone, it is possible to use OpenNebula's API and data structures to retrieve this ID based on the name of the desired VM or its IP address.
• Operating system & bootloader
  There are various ways to get to know a VMs OS, apart from simply guessing. For example, if the VM runs a publicly accessible web server, the OS of the VM could be leaked in the HTTP-Header Server (see RFC 2616). Another option would be to check the images or the template the VM was created from. Usually, the name and description of an image contains information about the installed OS, especially if the image was imported from a marketplace.
  Since most operating systems are shipped with a default bootloader, making a correct guess about a VMs bootloader is feasible. Even if this is not possible, other approaches can be used (see below).
• Keyboard layout of the VM's operating system
  As with the VMs bootloader, making an educated guess about a VM's keyboard layout is not difficult. For example, it is highly likely that VMs in a company's cloud will use the keyboard layout of the country the company is located in.

Overview of the Attack

The key idea of this attack is that neither Sunstone nor noVNC check whether keyboard related events were caused by human input or if they were generated by a script. This can be exploited so that gaining root access to a VM in OpenNebula requires five steps:

1. Using CSRF, a persistent XSS payload is deployed.
2. The XSS payload controls Sunstone's API.
3. The noVNC window of the VM to attack is loaded into an iFrame.
4. The VM is restarted using Sunstone's API.
5. Keystroke-events are simulated in the iFrame to let the bootloader open a root shell.

Figure 1: OpenNebula's Sunstone Interface displaying the terminal of a VM in a noVNC window.

The following sections give detailed information about each step.

Executing Remote Code in Sunstone

In Sunstone, every account can choose a display language. This choice is stored as an account parameter (e.g. for English LANG=en_US). In Sunstone, the value of the LANG parameter is used to construct a <script> tag that loads the corresponding localization script. For English, this creates the following tag:

<script src="locale/en_US/en_US.js?v=4.6.1" type="text/javascript"></script>

Setting the LANG parameter to a different string directly manipulates the path in the script tag. This poses an XSS vulnerability. By setting the LANG parameter to LANG="onerror=alert(1)//, the resulting script tag looks as follows:

<script src="locale/"onerror=alert(1)///"onerror=alert(1)//.js?v=4.6.1" type="text/javascript"></script>

For the web browser, this is a command to fetch the script locale/ from the server. However, this URL points to a folder, not a script. Therefore, what the server returns is no JavaScript. For the browser, this is an error, so the browser executes the JavaScript in the onerror statement: alert(1). The rest of the line (including the second alert(1)) is treated as comment due to the forward slashes.

When a user updates the language setting, the browser sends an XMLHttpRequest of the form

{ "action" : { "perform" : "update", "params" : { "template_raw" : "LANG=\"en_US\"" } }}

to the server (The original request contains more parameters. Since these parameters are irrelevant for the technique, we omitted them for readability.). Forging a request to Sunstone from some other web page via the victim's browser requires a trick since one cannot use an XMLHttpRequest due to restrictions enforced by the browser's Same-Origin-Policy. Nevertheless, using a self-submitting HTML form, the attacker can let the victim's browser issue a POST request that is similar enough to an XMLHttpRequest so that the server accepts it.

An HTML form field like

<input name='deliver' value='attacker' />

is translated to a request in the form of deliver=attacker. To create a request changing the user's language setting to en_US, the HTML form has to look like

<input name='{"action":{"perform":"update","params":{"template_raw":"LANG' value='\"en_US\""}}}' />

Notice that the equals sign in LANG=\"en_US\" is inserted by the browser because of the name=value format.

Figure 2: OpenNebula's Sunstone Interface displaying a user's attributes with the malicious payload in the LANG attribute.

Using this trick, the attacker sets the LANG parameter for the victim's account to "onerror=[remote code]//, where [remote code] is the attacker's exploit code. The attacker can either insert the complete exploit code into this parameter (there is no length limitation) or include code from a server under the attacker's control. Once the user reloads Sunstone, the server delivers HTML code to the client that executes the attacker's exploit.

Prepare Attack on VM

Due to the overwritten language parameter, the victim's browser does not load the localization script that is required for Sunstone to work. Therefore, the attacker achieved code execution, but Sunstone breaks and does not work anymore. For this reason, the attacker needs to set the language back to a working value (e.g. en_US) and reload the page in an iFrame. This way Sunstone is working again in the iFrame, but the attacker can control the iFrame from the outside. In addition, the attack code needs to disable a watchdog timer outside the iFrame that checks whether Sunstone is correctly initialized.

From this point on, the attacker can use the Sunstone API with the privileges of the victim. This way, the attacker can gather all required information like OpenNebula's internal VM ID and the keyboard layout of the VM's operating system from Sunstone's data-structures based on the name or the IP address of the desired VM.

Compromising a VM

Using the Sunstone API the attacker can issue a command to open a VNC connection. However, this command calls window.open, which opens a new browser window that the attacker cannot control. To circumvent this restriction, the attacker can overwrite window.open with a function that creates an iFrame under the attacker's control.

Once the noVNC-iFrame has loaded, the attacker can send keystrokes to the VM using the dispatchEvent function. Keystrokes on character keys can be simulated using keypress events. Keystrokes on special keys (Enter, Tab, etc.) have to be simulated using pairs of keydown and keyup events since noVNC filters keypress events on special keys.

Getting Root Access to VM

To get root access to a VM the attacker can reboot a victim's VM using the Sunstone API and then control the VM's bootloader by interrupting it with keystrokes. Once the attacker can inject commands into the bootloader, it is possible to use recovery options or the single user mode of Linux based operating systems to get a shell with root privileges. The hardest part with this attack is to get the timing right. Usually, one only has a few seconds to interrupt a bootloader. However, if the attacker uses the hard reboot feature, which instantly resets the VM without shutting it down gracefully, the time between the reboot command and the interrupting keystroke can be roughly estimated.

Even if the bootloader is unknown, it is possible to use a try-and-error approach. Since the variety of bootloaders is small, one can try for one particular bootloader and reset the machine if the attack was unsuccessful. Alternatively, one can capture a screenshot of the noVNC canvas of the VM a few seconds after resetting the VM and determine the bootloader.

A video of the attack can be seen here. The browser on the right hand side shows the victim's actions. A second browser on the left hand side shows what is happening in OpenNebula. The console window on the bottom right shows that there is no user-made keyboard input while the attack is happening.

   If you're a Windows user, follow these steps:

• First, you must download and run Ruby-lang setup file from RubyInstaller.org, choose Add Ruby executables to your PATH and Use UTF-8 as default external encoding.
• Then, download and install curl (32-bit or 64-bit) from Curl.haxx.se/windows. After that, go to Nmap.org/download.html to download and install the lastest Nmap version.
• Download killshot-master.zip and unzip it.
• Open CMD or PowerShell window at the KillShot folder you've just unzipped and enter these commands:
  ruby setup.rb
ruby killshot.rb

KillShot usage examples

   Easy and fast use of KillShot:

   Use KillShot to detect and scan CMS vulnerabilities (Joomla and WordPress) and scan for XSS and SQL:


References: Vulnrabilities are taken from

• Exploit Database - Exploits for Penetration Testers, Researchers, and Ethical Hackers
• Sploitus | Exploit & Hacktool Search Engine
• CXSECURITY.COM Free Security List

Download KillShot

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HTML5 Games On Android

On my last hollidays, I made two HTML5 games, and published on android market. Nowadays javascript has powerful libraries for doing almost everything, and also there are several compilers from java or c code to javascript, converting opengl c code to html5 canvas, but definitely, javascript execution is slower than dalvik applications, and of course much slower than arm c libs. For improving the speed of sounds and images loader, I have used javascript asynchronous execution and scheduling priority has been controlled with setTimeout/setInterval which deprioritize or priorize a code block. This games are published on the android market here: Android Planets and here: Far Planet

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Remot3d - An Easy Way To Exploiting

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SANS SEC575 Mentor Class

Hi everyone,

Great news! I will be mentoring SANS 575: Mobile Device Security and Ethical Hacking in Luxembourg on Thursday evenings 18:00-20:00, starting from January 15, 2015.

Mentor classes are special, 10 week-format SANS classroom sessions that give the students time to absorb and master the same material with the guidance of a trained security professional.

Students receive all the same course materials used at SANS conferences and study at a more leisurely pace, so students will have:

• Hardcopy set of SANS course books
• Mentor Program study materials
• Weekly Mentor led sessions

Prior to the weekly Mentor-led classroom sessions, students study SANS course material at their own pace. Each week, students meet with other professionals in their hometown area and the SANS mentor, who leads topical discussions pointing out the most salient features of the weekly material studied, provides hands-on demonstrations, and answer questions. The Mentor's goal is to help student's grasp the more difficult material, master the exercises, demonstrate the tools and prepare for GIAC certification.

On SANS SEC575, we will learn about mobile device infrastructures, policies and management, we will see the security models of the different platforms, like the data storage and file system architecture. We will also see how to unlock, root and jailbreak mobile devices in order to prepare them for data extraction and further testing. In the second half of the course, we will learn how to perform static and dynamic mobile application analysis, the usage of automated application analysis tools and how to manipulate application behavior. Last but not least, we will see how to perform mobile penetration testing that includes fingerprinting mobile devices, wireless network probing and scanning, attacking wireless infrastructures, using network manipulation attacks and attacks against mobile applications and back-end applications.

For more info, here is the link for the class: http://www.sans.org/mentor/class/sec575-luxembourg-15jan2015-david-szili
My Mentor bio: http://www.sans.org/mentor/bios#david-szili 

Information on the class, special discounts and applying for the class: szili_(dot)_david_(at)_hotmail_(dot)_com

Additional info can be also found at: https://www.sans.org/mentor

Some special price is also available for this course. A few examples: http://www.sans.org/mentor/specials

Best regards,

David

Such low price. Very SANS. Much learning. Wow!

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Hacktivity 2018 Badge - Quick Start Guide For Beginners

You either landed on this blog post because 

• you are a huge fan of Hacktivity
• you bought this badge around a year ago
• you are just interested in hacker conference badge hacking. 

or maybe all of the above. Whatever the reasons, this guide should be helpful for those who never had any real-life experience with these little gadgets. 

But first things first, here is a list what you need for hacking the badge:

• a computer with USB port and macOS, Linux or Windows. You can use other OS as well, but this guide covers these
• USB mini cable to connect the badge to the computer
• the Hacktivity badge from 2018

By default, this is how your badge looks like.

Let's get started

Luckily, you don't need any soldering skills for the first steps. Just connect the USB mini port to the bottom left connector on the badge, connect the other part of the USB cable to your computer, and within some seconds you will be able to see that the lights on your badge are blinking. So far so good. 

Now, depending on which OS you use, you should choose your destiny here.

Linux

The best source of information about a new device being connected is

# dmesg

The tail of the output should look like

[267300.206966] usb 2-2.2: new full-speed USB device number 14 using uhci_hcd
[267300.326484] usb 2-2.2: New USB device found, idVendor=0403, idProduct=6001
[267300.326486] usb 2-2.2: New USB device strings: Mfr=1, Product=2, SerialNumber=3
[267300.326487] usb 2-2.2: Product: FT232R USB UART
[267300.326488] usb 2-2.2: Manufacturer: FTDI
[267300.326489] usb 2-2.2: SerialNumber: AC01U4XN
[267300.558684] usbcore: registered new interface driver usbserial_generic
[267300.558692] usbserial: USB Serial support registered for generic
[267300.639673] usbcore: registered new interface driver ftdi_sio
[267300.639684] usbserial: USB Serial support registered for FTDI USB Serial Device
[267300.639713] ftdi_sio 2-2.2:1.0: FTDI USB Serial Device converter detected
[267300.639741] usb 2-2.2: Detected FT232RL
[267300.643235] usb 2-2.2: FTDI USB Serial Device converter now attached to ttyUSB0

Dmesg is pretty kind to us, as it even notifies us that the device is now attached to ttyUSB0. 

From now on, connecting to the device is exactly the same as it is in the macOS section, so please find the "Linux users, read it from here" section below. 

macOS

There are multiple commands you can type into Terminal to get an idea about what you are looking at. One command is:

# ioreg -p IOUSB -w0 -l

With this command, you should get output similar to this:

+-o FT232R USB UART@14100000  <class AppleUSBDevice, id 0x100005465, registered, matched, active, busy 0 (712 ms), retain 20>
    |   {
    |     "sessionID" = 71217335583342
    |     "iManufacturer" = 1
    |     "bNumConfigurations" = 1
    |     "idProduct" = 24577
    |     "bcdDevice" = 1536
    |     "Bus Power Available" = 250
    |     "USB Address" = 2
    |     "bMaxPacketSize0" = 8
    |     "iProduct" = 2
    |     "iSerialNumber" = 3
    |     "bDeviceClass" = 0
    |     "Built-In" = No
    |     "locationID" = 336592896
    |     "bDeviceSubClass" = 0
    |     "bcdUSB" = 512
    |     "USB Product Name" = "FT232R USB UART"
    |     "PortNum" = 1
    |     "non-removable" = "no"
    |     "IOCFPlugInTypes" = {"9dc7b780-9ec0-11d4-a54f-000a27052861"="IOUSBFamily.kext/Contents/PlugIns/IOUSBLib.bundle"}
    |     "bDeviceProtocol" = 0
    |     "IOUserClientClass" = "IOUSBDeviceUserClientV2"
    |     "IOPowerManagement" = {"DevicePowerState"=0,"CurrentPowerState"=3,"CapabilityFlags"=65536,"MaxPowerState"=4,"DriverPowerState"=3}
    |     "kUSBCurrentConfiguration" = 1
    |     "Device Speed" = 1
    |     "USB Vendor Name" = "FTDI"
    |     "idVendor" = 1027
    |     "IOGeneralInterest" = "IOCommand is not serializable"
    |     "USB Serial Number" = "AC01U4XN"
    |     "IOClassNameOverride" = "IOUSBDevice"
    |   } 

The most important information you get is the USB serial number - AC01U4XN in my case.
Another way to get this information is

# system_profiler SPUSBDataType

which will give back something similar to:

FT232R USB UART:

          Product ID: 0x6001
          Vendor ID: 0x0403  (Future Technology Devices International Limited)
          Version: 6.00
          Serial Number: AC01U4XN
          Speed: Up to 12 Mb/sec
          Manufacturer: FTDI
          Location ID: 0x14100000 / 2
          Current Available (mA): 500
          Current Required (mA): 90
          Extra Operating Current (mA): 0

The serial number you got is the same.

What you are trying to achieve here is to connect to the device, but in order to connect to it, you have to know where the device in the /dev folder is mapped to. A quick and dirty solution is to list all devices under /dev when the device is disconnected, once when it is connected, and diff the outputs. For example, the following should do the job:

ls -lha /dev/tty* > plugged.txt
ls -lha /dev/tty* > np.txt
vimdiff plugged.txt np.txt

The result should be obvious, /dev/tty.usbserial-AC01U4XN is the new device in case macOS. In the case of Linux, it was /dev/ttyUSB0.

Linux users, read it from here. macOS users, please continue reading

Now you can use either the built-in screen command or minicom to get data out from the badge. Usually, you need three information in order to communicate with a badge. Path on /dev (you already got that), speed in baud, and the async config parameters. Either you can guess the speed or you can Google that for the specific device. Standard baud rates include 110, 300, 600, 1200, 2400, 4800, 9600, 14400, 19200, 38400, 57600, 115200, 128000 and 256000 bits per second. I usually found 1200, 9600 and 115200 a common choice, but that is just me.
Regarding the async config parameters, the default is that 8 bits are used, there is no parity bit, and 1 stop bit is used. The short abbreviation for this is 8n1. In the next example, you will use the screen command. By default, it uses 8n1, but it is called cs8 to confuse the beginners.

If you type:
# screen /dev/tty.usbserial-AC01U4XN 9600
or
# screen /dev/ttyUSB0 9600
and wait for minutes and nothing happens, it is because the badge already tried to communicate via the USB port, but no-one was listening there. Disconnect the badge from the computer, connect again, and type the screen command above to connect. If you are quick enough you can see that the amber LED will stop blinking and your screen command is greeted with some interesting information. By quick enough I mean ˜90 seconds, as it takes the device 1.5 minutes to boot the OS and the CTF app.

Windows

When you connect the device to Windows, you will be greeted with a pop-up.

Just click on the popup and you will see the COM port number the device is connected to:

In this case, it is connected to COM3. So let's fire up our favorite putty.exe, select Serial, choose COM3, add speed 9600, and you are ready to go!

You might check the end of the macOS section in case you can't see anything. Timing is everything.

The CTF

Welcome to the Hacktivity 2018 badge challenge!

This challenge consists of several tasks with one or more levels of
difficulty. They are all connected in some way or another to HW RE
and there's no competition, the whole purpose is to learn things.

Note: we recommend turning on local echo in your terminal!
Also, feel free to ask for hints at the Hackcenter!

Choose your destiny below:

  1. Visual HW debugging
  2. Reverse engineering
  3. RF hacking
  4. Crypto protection

Enter the number of the challenge you're interested in and press [

Excellent, now you are ready to hack this! In case you are lost in controlling the screen command, go to https://linuxize.com/post/how-to-use-linux-screen/.

I will not spoil any fun in giving out the challenge solutions here. It is still your task to find solutions for these.

But here is a catch. You can get a root shell on the device. And it is pretty straightforward. Just carefully remove the Omega shield from the badge. Now you see two jumpers; by default, these are connected together as UART1. As seen below.

But what happens if you move these jumpers to UART0? Guess what, you can get a root shell! This is what I call privilege escalation on the HW level :) But first, let's connect the Omega shield back. Also, for added fun, this new interface speaks on 115200 baud, so you should change your screen parameters to 115200. Also, the new interface has a different ID under /dev, but I am sure you can figure this out from now on.

If you connect to the device during boot time, you can see a lot of exciting debug information about the device. And after it boots, you just get a root prompt. Woohoo! 

But what can you do with this root access? Well, for starters, how about running 

# strings hello | less

From now on, you are on your own to hack this badge. Happy hacking.

Big thanks to Attila Marosi-Bauer and Hackerspace Budapest for developing this badge and the contests.

PS: In case you want to use the radio functionality of the badge, see below how you should solder the parts to it. By default, you can process slow speed radio frequency signals on GPIO19. But for higher transfer speeds, you should wire the RF module DATA OUT pin with the RX1 free together.

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BeEF: Browser Exploitation Framework

"BeEF is the browser exploitation framework. A professional tool to demonstrate the real-time impact of XSS browser vulnerabilities. Development has focused on creating a modular structure making new module development a trivial process with the intelligence residing within BeEF. Current modules include the first public Inter-protocol Exploit, a traditional browser overflow exploit, port scanning, keylogging, clipboard theft and more." read more...


Website: http://www.bindshell.net/tools/beef

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