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 = list(input())
  
  if s[0] == "k":
      li = []
      for i in range(6):
          li.append(s[len(s) - (6 - i)])
      if "".join(li) == "eyence":
          print("YES")
          quit()
  
  if s[0] + s[1] == "ke":
      li = []
      for i in range(5):
          li.append(s[len(s) - (5 - i)])
      if "".join(li) == "yence":
          print("YES")
          quit()
  
  if s[0] + s[1] + s[2] == "key":
      li = []
      for i in range(4):
          li.append(s[len(s) - (4 - i)])
      if "".join(li) == "ence":
          print("YES")
          quit()
  
  if s[0] + s[1] + s[2] + s[3] == "keye":
      li = []
      for i in range(3):
          li.append(s[len(s) - (3 - i)])
      if "".join(li) == "nce":
          print("YES")
          quit()
  
  if s[0] + s[1] + s[2] + s[3] + s[4] == "keyen":
      li = []
      for i in range(2):
          li.append(s[len(s) - (2 - i)])
      if "".join(li) == "ce":
          print("YES")
          quit()
  
  if s[0] + s[1] + s[2] + s[3] + s[4] + s[5] == "keyenc":
      if s[len(s) - 1] == "e":
          print("YES")
          quit()
  
  
  lis = []
  for i in range(7):
      lis.append(s[len(s) - i - 1])
  if "".join(lis) == "keyence":
      print("YES")
      quit()
  
  print("NO")
  
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"}]