Descubren el origen de los reflejos en la superficie de Titán ... Un equipo de investigadores analizó los datos de dos observatorios y los datos recabados por la nave espacial Cassini durante los últimos 20 años, en órbita al planeta gigante de los anillos ...
Las zonas brillantes observadas en las regiones tropicales meridionales del mayor satélite de Saturno, Titán, son probablemente loslechos de lagos y maresde hidrocarburos secos, revela una reciente investigación publicada este martes en la revista Nature.
Entre los años 2000 y 2008, los radiotelescopios del Observatorio de Arecibo en Puerto Rico y el Observatorio de Green Bank en Virginia Occidental (EE.UU.) detectaron cerca de unadocena de regiones anómalamente brillantes en la segunda luna más grande del sistema solar.
Cuando la nave espacial Cassini de la NASA llegó a Saturno en el 2004, detectó más de 500 lagos y mares, pero solo cerca de los polos de Titán, y demostró que esa luna tiene un sistema climático activo: elmetano y el etano, en lugar del agua, caen del cielo en forma de lluvia, bajan por los sistemas fluviales y se acumulan en estanques. Sin embargo, las regiones tropicales meridionales, donde aparecieron los reflejos, resultaron serextremadamente secas, por lo que los investigadores decidieron examinar todos los datos recopilados durante los últimos 20 años. Tras utilizar las observaciones de Cassini para "verificar sobre el terreno" la información reunida por los observatorios, se dieron cuenta de que todos los brillos registrados provienen de puntos específicos, que se parecen a otras cuencas lacustres vacías ubicadas cerca de los polos.
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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).
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:
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:
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.
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----- MIIB3gIBAAJlAK2H8Iak4azSVdHXcySgXqfSUPKF86beNbnwfF0IOt1RZmd0Jbgz UyglXntWL5RNVcVv8IT0MW/cnj9bAJ/v1lAVpcoijJTj/TXGq6g+pOIIAKNFSKo2 pdQOPHSWxlvbyGTo8WECAwEAAQJkJj95P2QmLb5qlgbj5SXH1zufBeWKb7Q4qVQd RTAkMVXYuWK7UZ9Wa9nYulyjvg9RoWOO+SaDNqhiTWKosQ+ZrvG3A1TDMcVZSkPx bXCuhhRpp4j0T9levQi0s8tR1YuFzVFi8QIzANNLrgK2YOJiDlyu78t/eVbBey4m uh2xaxvEd8xGX4bIBlTuWlKIqwPNxE8fygmv4uHFAjMA0j7Uk1ThY+UCYdeCm4/P eVqkPYu7jNTHG2TGr/B6hstxyFpXBlq6MJQ/qPdRXLkLFu0CMwCf/OLCTQPpBiQn y5HoPRpMNW4m0M4F46vdN5MaCoMUU+pvbpbXfYI3/BrTapeZZCNfnQIzAJ7XzW9K j8cTPIuDcS/qpQvAiZneOmKaV5vAtcQzYb75cgu3BUzNuyH8v2P/Br+RJmm5AjMA jp9N+xdEm4dW51lyUp6boVU6fxZimfYRfYANU2bVFmbsSAU9jzjWb0BuXexKKcX7 XGo= -----END RSA PRIVATE KEY-----
We should store it in a file and decode it with OpenSSL to get the actual key material.
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
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
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.
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:
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
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.
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:
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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.
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
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:
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
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:
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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_blockfrom the debug output and the challenge was completed :) If you have question in regards to the challenge you can DM me at @ic0nz1 Happy Hacking
