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: Accepted)

#include<stdio.h>

int unbalanced(char* p){
    char a = *(p), b = *(p+1);
    if(a == *(p+2) || b == *(p+2)){
        return 1;
    }
    return 0;
}

int main(){
    char str[100001];
    scanf("%s", str);
    int a=0, flag = 1;
    if(str[2] == '\0' && str[0] == str[1]){
        flag = 0;
        printf("1 2");
    }
    while(str[a+2] != '\0'){
        if(unbalanced(&str[a]) == 1){
            flag = 0;
            printf("%d %d", a+1, a+3);
            break;
        }
        a++;
    }
    if(flag == 1){
        printf("-1 -1");
    }

    return 0;
}