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;

#define FOR(i, n) for(int i = 0; i < N; i++)

void cin_Ignore(){cin.ignore(numeric_limits<streamsize>::max(), '\n');}

int main(){
    
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    string input;
    
    cin >> input; cin_Ignore;
    
    unordered_map<char, vector<long> > Area;
    
    for(int i = 0; input[i]; i++){
        Area[input[i]].push_back(i+1);
    }
    
    unordered_map<char, long> Area_size;
    for(const auto &fd : Area){
        Area_size[fd.first] = fd.second.size();
    }
    
    
    pair<char, long> k;
    k = *max_element(Area_size.begin(), Area_size.end(), [](const auto &fd, const auto &fe){
        return fd.second < fe.second;
    });
    
    if(k.second > input.length()/2){
        cout << *Area[k.first].begin() << " " << *(Area[k.first].end()-1);
    }
    else
        cout << "-1 -1";
    
}