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>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <cmath>

using namespace std;


bool is_balanced(int s, int e, vector<int>& v) {
    int len = e - s + 1;
    for (auto i: v) {
        if (i > len / 2 && len > 2) {
            return true;
        }
    }
    return false;
}

void print(vector<int> v) {
    cout << "{ ";
    for (auto i : v) {
        cout << i << ", ";
    }
    cout << " }" << endl;
}

int main() {
    string s;
    cin >> s;
    vector<int> v(26, 0);

    int i = 0, j = 0;
    int len = (int) s.size();
    int start, end;

    while (j < len) {
        v[s[j] - 'a']++;
        if (is_balanced(i, j, v)) {
            start = i + 1;
            end = j + 1;
            cout << start << " " << end << endl;
            return 0;
        } else {
            if (j > i + 1) {
                v[s[i] - 'a']--;
                i++;
            }
        }
        j++;
    }

    cout << -1 << " " << -1 << endl;

    return 0;
}