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: Wrong Answer)

import java.util.Scanner;
import java.util.HashMap;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a string: ");
        String string = sc.nextLine();

        boolean found = false;
        int currLen = 2;
        int currIdx;
        HashMap<Character, Integer> map;

        while (currLen < string.length() + 1 && !found) {
            currIdx = 0;
            while (currIdx + currLen - 1 < string.length()) {
                map = new HashMap<>();
                for (int i = 0; i < currLen; i++) {
                    char currChar = string.charAt(i + currIdx);
                    if (map.containsKey(currChar)) {
                        map.replace(currChar, map.get(currChar) + 1);
                    } else {
                        map.put(currChar, 1);
                    }
                }
                boolean unbalance = false;
                for (Integer value: map.values()) {
                    if (value > currLen / 2) {
                        unbalance = true;
                    }
                }
                //System.out.println((currLen));
                if (unbalance) {
                    System.out.print(currIdx + 1 + " " + (currIdx + currLen));
                    System.out.println();
                    found = true;
                }
                currIdx++;
            }
            currLen++;
        }

        if (!found) {
            System.out.println(-1 + " " + -1);
        }
    }
}