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 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);

    string s;
    cin>>s;
    vector<pair<string,pair<int,int>>>v;
    for(int i=0;i<s.size();i++)
    {

        string x="";
        for(int j=i;j<s.size();j++)
        {
            x+=s[j];
            v.push_back({x,{i,j}});
        }
    }

        for(int i=0;i<v.size();i++)
        {
            string s1=v[i].first;
            int ind1=v[i].second.first;
            int ind2=v[i].second.second;
            map<char,int>mp;
            if(s1.size()>1)
            {
                int len=(s1.size()+2)/2;
                for(int k=0;k<s1.size();k++)
                {
                    mp[s1[k]]++;
                    if(mp[s1[k]]>=len)
                    {
                       //cout<<s1<<"\n";
                        cout<<ind1+1<<" "<<ind2+1<<"\n";
                        return 0;
                    }
                }
            }
        }

    cout<<-1<<" "<<-1<<"\n";


}