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: Runtime Error)

#include <stdio.h>
#include <string.h>
int che_pat(char* s,char c,int* a,int* b) {
	int f[2] = { 0,0 };
	for (int i = 0;s[i] != '\0';i++) {
		for (int ii = i + 1;s[ii] != '\0';ii++) {
			if (s[i] == c) {
				if (s[i] == s[ii]) {   //xx
					f[0] = 1;
					*a = i;
				}
				if (s[i] == s[i + 2]) { //xyx
					f[1] = 1;
					*b = ii;
				}
			}
		}
			if (f[0] == 1 && f[1] == 1) {
				return 1;
				break;
            }
	}
	return 0;
}
int ch_majority(char* s, char* c) {
	char  ch;
	int num = 0, j = 0;
	//アルファベット順に並び替える
	for (int i = 0;s[i] != '\0';i++) {
		for (int ii = i + 1;s[ii] != '\0';ii++) {
			if (s[i] > s[ii]) {
				ch = s[i];
				s[i] = s[ii];
				s[ii] = ch;
			}
		}
	}
	//カウント  //cが過半数の文字
	while (1) {
		if (num == 0) *c = s[j];
		if (*c == s[j]) num++;
		else num--;
		j++;
		if (s[j] == '\0') break;
	}
	return 0;
}
int main() {
	char s[100], c, st[100];
	int num = 0;
	int cn = 0;
	int a = 0, b = 0;
	scanf("%s", s, 100);
	strcpy(st, s);
	ch_majority(&s[0], &c);
	if (che_pat(&st[0], c,&a,&b)) printf("%d %d\n", a, b);
	else printf("-1 -1\n");
	return 0;
}