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 (Clang 3.8.0) (judge verdict: Runtime Error)

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

int main(int argc, char const *argv[]) {
  int i, j, length, cnt;
  char lead;
  char str[10001];
  scanf("%s", str);
  // printf("%s %lu", str, strlen(str));
  for(length=3; length<strlen(str); length+=2){
    for(i=0; i<strlen(str)-length+1; i++){
      cnt = 0;
      lead = str[i];
      // printf("lead %d %c", i, lead);
      for(j=i; j<i+length; j++){
        // printf("check %d %c", j, str[j]);
        if(str[j]==lead)
          cnt++;
      }
      if(cnt>length/2){
        printf("%d %d\n", i+1, i+length);
        return 0;
      }
    }
  }
  printf("-1 -1\n");
  return 0;
}