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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner in=new Scanner (System.in);
		String s=in.next();
		int a=-1;
		int b=-1;
		int c=0;
		int subL=0;
		for (int i = 0; i < s.length(); i++) {
			for (int j = 0; j < s.length(); j++) {
				if(s.charAt(i)==s.charAt(j) && i!=j) {
					c++;
					a=i+1;
					b=j+1;
				}
				subL=b-a+1;
				if(subL>=2 && c>(double)subL/2) {
					System.out.println(a+" "+b);
					c=1000000;
					break;
				}
			
			}
			if(c==1000000)
				break;
			c=0;
			if(i==s.length()-1 && c==0)
				System.out.println(-1+" "+-1);
		}
	}
}