Problem:
Score : 400 points

Problem 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.

Constraints

2 ≦ |s| ≦ 10^5

s consists of lowercase letters.

Partial Score

200 points will be awarded for passing the test set satisfying 2 ≦ N ≦ 100.

Input

The input is given from Standard Input in the following format:

s

Output

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.

Sample Input 1

needed

Sample Output 1

2 5

The string s_2 s_3 s_4 s_5 = eede is unbalanced. There are also other unbalanced substrings. For example, the output 2 6 will also be accepted.

Sample Input 2

atcoder

Sample Output 2

-1 -1

The string atcoder contains no unbalanced substring.

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

s = input()
dic = {}
flag = False

for key in s:
  if key not in dic:
    dic[key] = 0
  dic[key] += 1
#print(dic)


if len(s) % 2 == 0:
#  print('yes')
  chk = 0
  counter = 0
  for k, v in dic.items():
    counter += 1
    if v == len(s)/2:
      chk += 1
      continue
    if chk >= 2:
      break
    elif chk < 2 and counter == len(dic)-1:
      for k, v in dic.items():
        if v == len(s)/2:
          start = s.find(k)+1
          end = len(s)
          print(start, end)
          flag = True
          break
    if v > len(s)/2:
      start = s.find(k)+1
      end = len(s)
      print(start, end)
      flag = True
      break
      
elif len(s) % 2 == 1:
  for k, v in dic.items():
    if v >= len(s)/2:
      start = s.find(k)+1
      end = len(s)
      print(start, end)
      flag = True
      break

if flag == False:
  print(-1, -1)