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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

/* Filename: ARC059D.java
 * Author: Mushiyo
 */

import java.util.Scanner;
import java.util.Arrays;

public class Main{

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);

		while (input.hasNext()) {
			String s = input.next();
			int[] countChar = new int[26];
			int[] countPrevFreq = new int[s.length()];
			int[] prevPos = new int[26];
			Arrays.fill(prevPos, -1);

			int a = -1, b = -1;
			for (int i = 0; i < s.length(); ++i) {
				int currentChar = s.charAt(i) - 'a';
				countPrevFreq[i] = countChar[currentChar];
				++countChar[currentChar];

				if (prevPos[currentChar] > -1) {
					int len = i - prevPos[currentChar] + 1;
					int freqDiff = countPrevFreq[i]
							- countPrevFreq[prevPos[currentChar]] + 1;

					if (freqDiff * 2 > len) {
						a = prevPos[currentChar] + 1;
						b = i + 1;
					}

				}

				prevPos[currentChar] = i;
			}

			System.out.printf("%d %d\n", a, b);
		}
	}
}