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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int c[26];

int main()
{
    string a;
    int b;
    char d[26],e='a';
    for(int i=0; i<26; i++)
    {
        d[i]=e;
        e++;
    }
    cin>>a;
    for(int i=0; i<a.size(); i++)
    {
        for(int j=0; j<26; j++)
        {
            if(a[i]==d[j])
            {
                c[j]++;
                break;
            }
        }
    }
    vector<int>v;
    for(int i=0; i<26; i++)
    {
        if(c[i]>0)
        {
            v.push_back(c[i]);
        }
    }
    sort(v.begin(),v.end());
    if(v[0]==v[v.size()-1])
    {
        cout<<"-1 -1"<<endl;
        return 0;
    }
    else
    {
        for(int i=0; i<a.size()-2; i++)
        {
            if(a[i]==a[i+1])
            {
                cout<<i+1<<" "<<i+1+1<<endl;
                break;
            }
            if(a[i]==a[i+2])
            {
                cout<<i+1<<" "<<i+2+1<<endl;
                break;
            }
        }
    }
}