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[100000],p;
    int i,l,a,z;
    while(scanf("%s",s)!=EOF){
    int c[26]={0},i=0;
    l=strlen(s);
    while(s[i]!='\0'){
       if(s[i]>='a'&&s[i]<='z')
       c[s[i]-'a']++;
       i++;
    }
    a=c[0];
    for (i=1;i<26;i++){
       if(a<c[i]){a=c[i];p=i+'a';}
    }
    if(2*a>=l){
        for(i=0;i<l;i++){
            if(p==s[i]){
                printf("%d ",i+1);
                goto mhe;
            }
        }
        mhe:
        for(i=l-1;i>=0;i--){
            if(p==s[i]){printf("%d\n",i+1);break;}
             }
         }
    else printf("-1 -1\n");
   }
   return 0;
}