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: Accepted)

#include <bits/stdc++.h>

using namespace std;

namespace {

    typedef double real;
    typedef long long ll;

    template<class T> ostream& operator<<(ostream& os, const vector<T>& vs) {
        if (vs.empty()) return os << "[]";
        os << "[" << vs[0];
        for (int i = 1; i < vs.size(); i++) os << " " << vs[i];
        return os << "]";
    }
    template<class T> istream& operator>>(istream& is, vector<T>& vs) {
        for (auto it = vs.begin(); it != vs.end(); it++) is >> *it;
        return is;
    }

    string s;
    void input() {
        cin >> s;
    }

    bool check(int x, int y) {
        if (s.size() < y) return false;
        int n = y - x;
        map<char, int> count;
        for (int j = x; j < y; j++) {
            count[s[j]]++;
        }
        for (auto it = count.begin(); it != count.end(); it++) {
            if (it->second > double(n) / 2) return true;
        }
        return false;
    }

    void solve() {
        for (int i = 0; i < s.size(); i++) {
            if (check(i, i + 2)) {
                cout << i + 1 << " " << i + 2 << endl;
                return;
            }
            if (check(i, i + 3)) {
                cout << i + 1 << " " << i + 3 << endl;
                return;
            }
        }
        cout << "-1 -1" << endl;
    }
}

int main() {
    input(); solve();
    return 0;
}