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.Arrays;
import java.util.ArrayList;
import java.util.List;
import java.util.Collections;

public class Main {
	public static void main(String args[]){
		    Scanner sc = new Scanner(System.in);	
			
			String s = sc.next();
			String ss = s.substring(0,1);
			String tmp = s.replaceAll(ss,"a");
			String a = "";
			int pricnt = 0;

			for (int i = 0; i < s.length(); i++) {
				a = a + "a";
			}
			if (tmp.equals(a)) {
				System.out.println("-1 -1");
			}else{
				for (int i = 0; i < s.length()-1; i++) {
					if (i == 0) {
						if (s.charAt(i) == s.charAt(i+1) 
								&& s.charAt(i) != s.charAt(i+2)) {
							System.out.println(1 + " " + 3);
							pricnt++;
						}
					}else if(i != s.length()-1){
						if (s.charAt(i)==s.charAt(i+1)) {
							System.out.println(i + " " + (i+2));
							pricnt++;
						}
					}
					if (pricnt == 1) {
						break;
					}
				}
				if (pricnt == 0) {
					for (int i = 0; i < s.length()-2; i++) {
						if (s.charAt(i) == s.charAt(i+2)) {
							pricnt++;
							System.out.println((i+1) +" " + (i+3));
						}
						if (pricnt == 1) {
							break;
						}
					}
				}

				if (pricnt == 0) {
					System.out.println("-1 -1");
				}

			}


		    sc.close(); }

	public static boolean isEven(int n ) {
		if (n % 2 == 0) {
			return true;
		}else{
			return false;
		}
	}
}