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.io.*;
public class Main
{
	public static void main(String args[])
    {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		try {
			
		
			String str = br.readLine();
			String data[] = new String[str.length()];
			int data2[] = new int[str.length()];
			int flag = 0;
			
			if(str.length() < 2){
				flag=-1;
			}

			for(int i=0;i<str.length();i++){
				data[i]=str.substring(i,i+1);
				data2[i]=0;
			}
			
			for(int i=0;i<str.length();i++){
				for(int j=0;j<str.length();j++){
					if(data[i].equals(data[j])){
						data2[i]++;
					}
				}
			}
			
			int max = 0;
			int max_i = 0;
			for(int i=0;i<str.length();i++){
				if(max < data2[i]){
					max = data2[i];
					max_i = i;
				}
			}
			
			int start=str.length()+1, end=-1;
			for(int i=0;i<str.length();i++){
				if(start > i && data[i].equals(data[max_i])){
					start = i;
				}
				if(end < i && data[i].equals(data[max_i])){
					end = i;
				}
			}

			if(flag==-1 || max<=1){
				System.out.println("-1 -1");
			}else{
				start++;
				end++;
				System.out.println(start+" "+end);
			}


		} catch (Exception e) {
			// System.out.println(e);
        }
	}
}