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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int main() {
  char s[100005]={'\0'};
  int count[26]={0};
  char target='\0';
  int i=0,st=0,en=0;
  scanf("%s",s);
  while(s[i]!='\0'){
    count[((int)(s[i]-'a'))]++;
    i++;
  }
  for(int i=0;i<26;i++){
    if(count[i]>=strlen(s)/2){
      target= 'a' + i;
      break;
    }
  }
  for(int i=0;i<strlen(s);i++){
    if(s[i]==target){
      if(st==0) st=i+1;
      en=i+1;
    }
  }
  if((st==0)&&(en==0)){
    printf("-1 -1\n");
  }else{
    printf("%d %d\n",st,en);
  }
  return 0;
}