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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        char[] str = scan.next().toCharArray();
        int n = str.length;
        TreeMap<Double, Integer> sumIdx = new TreeMap<>();
        for (char c = 'a'; c <= 'z'; c++){
            double sum = 0;
            sumIdx.clear();
            sumIdx.put(0.0, -1);
            for (int i=0;i<n;i++) {
                sum += str[i] == c? 0.5: -0.5;
                Map.Entry<Double, Integer> lower = sumIdx.lowerEntry(sum);
                if (lower != null) {
                    if (lower.getValue() == i - 1) lower = sumIdx.lowerEntry(lower.getKey());
                    if (lower != null) {
                        System.out.println(lower.getValue() + 2 + " " + (i+1));
                        return;
                    }
                }
                sumIdx.putIfAbsent(sum, i);
            }
        }
        System.out.println("-1 -1");
    }
}