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 (Clang 3.8.0) (judge verdict: Accepted)

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

char str[100005];

int isUnb(char *p);
int f(char *p);

int main()
{
	int length;

	scanf("%s", str);

	length = strlen(str);
	for (int i = 0; i < length - 1; i++) {
		if (f(str + i)) {
			printf("%d %d\n", i + 1, i + 2);
			return 0;
		}
		if (isUnb(str + i)) {
			printf("%d %d\n", i + 1, i + 3);
			return 0;
		}
	}

	printf("-1 -1\n");

	return 0;
}

int isUnb(char *p)
{
	if (p[0] == p[1] || p[0] == p[2] || p[1] == p[2]) {
		return 1;
	} else {
		return 0;
	}
}

int f(char *p)
{
	if (p[0] == p[1]) {
		return 1;
	} else {
		return 0;
	}
}