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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>

int main(void)
{
	char str[1000000];
	
	scanf("%s", str);
	
	int i, j;
	
	int size = strlen(str);
	char ch;
	int n;
	
	for(i=0; i<size; i++)
	{
		ch = str[i];
		for(j=i, n=0; j<size; j++)
		{
			if(str[j]==ch)
			n++;
			
//			printf("%d", n);
			if(n>=((j-i+1)/2+1) && (j-i+1)>2)
			{
				printf("%d %d\n", i+1, j+1);
				return 0;
			}
		}
		
	}
	
	printf("-1 -1");
	
	
	return 0;
}