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>

char ch[100001];
char c[3];
int flag;
int a, b;

int main(void)
{
  int i;
  
  scanf("%s", ch);
  c[0] = ch[0];
  c[1] = ch[1];
  flag = (c[0] == c[1]);
  a = b = -1;
  for (i = 2; ch[i]; ++i)
  {
    c[i%3] = ch[i];
    if (flag && c[i%3] == c[(i+1)%3]) continue;
    if (flag == 0 && c[i%3] != c[(i+1)%3] && c[i%3] != c[(i+2)%3]) continue;
    a = i-2;
    b = i;
  }
  printf("%d %d\n", a, b);
  
  return 0;
}