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.io.*;
import java.util.*;
public class Main {
	
	public static void main(String[] args) {
		InputReader in = new InputReader();
		String str = in.next();
		
		for (int i=0; i<str.length()-1; i++) {
			if (str.charAt(i) == str.charAt(i+1)) {
				System.out.println((i+1) + " " + (i+2));
				return;
			}
			if (i < str.length()-2 && str.charAt(i) == str.charAt(i+2)) {
				System.out.println((i+1) + " " + (i+3));
				return;
			}
		}
		System.out.println(-1 + " " + -1);
	}

	static class InputReader {
		public BufferedReader br;
		public StringTokenizer st;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = null;
		}
		
		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return st.nextToken();
		}
	}
}