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

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

typedef char* String;

String string_input(void){
    int i,len;
    char buf[100001];
    String str;
    scanf("%s",buf);//buf=入力した文字列が入るchar型の配列
    
    len=strlen(buf);//len=文字列の長さを表すint型変数
    NEW(str,len+1);//strはただのポインタだから、ここでNEWを使ってメモリを確保しなくちゃいけない
    for(i=0;i<len;i++){
        str[i]=buf[i];
    }
    str[len]=0;
    
    return str;
}

int main(void){
    String S=string_input();
    int N=strlen(S);

    for(int i=0;i<N-2;i++){
        if(S[i]!=S[i+1] && S[i+1]!=S[i+2] && S[i]!=S[i+2]) continue;
        else{
            printf("%d %d\n",i+1,i+3);
            return 0;
        }
    }

    printf("-1 -1\n");
    return 0;
}