output_description: If there exists no unbalanced substring of s, print `-1 -1`.
  
  If there exists an unbalanced substring of s, let one such substring be s_a
  s_{a+1} ... s_{b} (1 ≦ a < b ≦ |s|), and print `a b`. If there exists more
  than one such substring, any of them will be accepted.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      s
attempt: #アンバランスな文字列からアンバランスでない文字列を取り除くと、
  #アンバランスな文字列が残るはず。ということは、両側から文字を削り取ることで「どの部分列をとってもアンバランスな文字列」が存在するはず
  #すなわちそれは、両側に「ax」が存在しない文字列で、「axa」または「aa...aa」
  #aaがある時点でアンバランス確定なので「axa」「aa」のどちらかを探せばよし
  import re
  s=input()
  alphabet="abcdefghijklmnopqrstuvwxyz"
  for i in alphabet:
      ba=s.find(i+i)
      if ba!=-1:
          print(str(ba+1)+str(ba+2))
          break
      A=re.compile(i+"."+i)
      axa=A.search(s,0)
      if axa:
          ba=axa.start())
          print(str(ba+1)+str(ba+3))
          break
problem_description: Statement
  
  Given a string t, we will call it _unbalanced_ if and only if the length of t
  is at least 2, and more than half of the letters in t are the same. For
  example, both `voodoo` and `melee` are unbalanced, while neither `noon` nor
  `a` is.
  
  You are given a string s consisting of lowercase letters. Determine if there
  exists a (contiguous) substring of s that is unbalanced. If the answer is
  positive, show a position where such a substring occurs in s.
samples: [{"input": "needed", "output": "2 5\n    \n\nThe string s_2 s_3 s_4 s_5 = `eede` is unbalanced. There are also other\nunbalanced substrings. For example, the output `2 6` will also be accepted.\n\n* * *"}, {"input": "atcoder", "output": "-1 -1\n    \n\nThe string `atcoder` contains no unbalanced substring."}]