output_description: If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      S
attempt: s = input()
  search="keyence"
  idx=0
  flg = False
  ans="NO"
  if s.find(search)>=0:
      print("YES")
  elif:
      for i in range(7):
          x=s.find(search[0:i+1])
          print("ps=%s, ns=%s, reduce=%s" %(search[0:i+1], search[i+1:], s[x+i+1:]))
          if x == -1:
              break
          if x == 0:
              if s[x+i:].find(search[i+1:])>=0:
                  ans = "YES"
                  break
      print(ans)
  
problem_description: Statement
  
  A string is called a KEYENCE string when it can be changed to `keyence` by
  removing its contiguous substring (possibly empty) only once.
  
  Given a string S consisting of lowercase English letters, determine if S is a
  KEYENCE string.
samples: [{"input": "keyofscience", "output": "YES\n    \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n    \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n    \n\n* * *"}, {"input": "keyence", "output": "YES"}]