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: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	private static final FastIn in = new FastIn();
	private static final PrintWriter out = new PrintWriter(System.out);
	public static void main(String[] args) {
		Solver sol = new Solver();
		sol.solve();
		out.close();
	}
	private static final class Solver {
		public void solve() {
			char[] s = in.next().toCharArray();
			int n = s.length;
			for (int i = 0; i < n - 1; ++i) {
				if (s[i] == s[i + 1]) {
					out.println((i + 1) + " " + (i + 2));
					return;
				}
			}
			out.println("-1 -1");
		}
	}
	private static final class FastIn {
		private BufferedReader br;
		private StringTokenizer st;
		public FastIn() {
			br = new BufferedReader(new InputStreamReader(System.in), 1 << 20);
			st = null;
		}
		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return st.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}