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 11.0.6) (judge verdict: Accepted)



import java.util.Scanner;

public class Main {

     static Scanner scanner;

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

         String S=gs();
         for (int i=0; i<S.length()-1;i++) {
        	 if(S.charAt(i)==S.charAt(i+1)) {
        		 System.out.println((i+1)+" "+(i+2));
        		 return;
        	 }
         }
         for (int i=0; i<S.length()-2;i++) {
        	 if(S.charAt(i)==S.charAt(i+2)) {
        		 System.out.println((i+1)+" "+(i+3));
        		 return;
        	 }
         }
         
         System.out.println("-1 -1");
     }

  // a^nをmodで割ったあまり
 	public static long modpow(long a, long n, long mod) {
 		long res=1;
 		while(n>0) {
 			if((n & 1) == 1) res=res*a%mod;
 			a=a*a%mod;
 			n=n>>1;
 		}

 		return res;
 	}

     // 文字列として入力を取得
     public static String gs() {
          return scanner.next();
     }

     // intとして入力を取得
     public static int gi() {
          return Integer.parseInt(scanner.next());
     }

     // longとして入力を取得
     public static long gl() {
          return Long.parseLong(scanner.next());
     }

     // doubleとして入力を取得
     public static double gd() {
          return Double.parseDouble(scanner.next());
     }
}