import requests
  import sys
  url = sys.argv[1]
  for i in open(sys.argv[2],"r"):
    #print i.rstrip() # removes newline character 
    r = requests.get("http://{}/{}".format(url,i.rstrip()))
    if r.status_code == 200:
        print "http://{}/{}".format(url,i.rstrip())
  
  ##to run
  python wfuzz_1.py 127.0.0.1 common.txt