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.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

	void run() {
		Scanner sc = new Scanner(System.in);

		char[] c = sc.next().toCharArray();
		int n = c.length;
		int size = 0;
		int s = 0;
		int t = 0;

		Map<Character, Integer> map = new HashMap<>();
		boolean f = false;
		boolean flag = true;

		// 尺取り法
		while (s <= t) {
			while (t < n) {
				char key = c[t];
				size = (t - s + 1) / 2;
				if (map.containsKey(key)) {
					int v = map.get(key);
					map.put(key, v + 1);
				} else {
					map.put(key, 1);
				}
				int v = map.get(key);
				if (v > size && size >= 1) {
					f = true;
					break;
				}
				t++;
			}
			if (f) {
				break;
			}
			char key = c[s];
			int v = map.get(key);
			map.put(key, v - 1);
			s++;
			size = (t - s + 1) / 2;
			if (s == n) {
				flag = false;
				break;
			}
			key = c[s];
			v = map.get(key);
			if (v > size) {
				break;
			}
		}

		if (flag) {
			if (t == n) {
				System.out.println((s + 1) + " " + t);
			} else {
				System.out.println((s + 1) + " " + (t + 1));
			}
		} else {
			System.out.println(-1 + " " + -1);
		}

	}

	public static void main(String[] args) {
		new Main().run();
	}

}