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.

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.

Examples

Input

needed


Output

2 5


Input

atcoder


Output

-1 -1
import java.util.*;
import java.io.*;
import java.awt.geom.*;
import java.math.*;

public class Main {

	static final Scanner in = new Scanner(System.in);
	static final PrintWriter out = new PrintWriter(System.out,false);
	static boolean debug = false;

	static void solve() {
		String s = in.next();
		int n = s.length();
		for (int i=0; i<n-1; i++) {
			if (s.charAt(i) == s.charAt(i+1)) {
				out.println((i+1)+" "+(i+2));
				return;
			}
		}
		for (int i=0; i<n-3; i++) {
			if (s.charAt(i) == s.charAt(i+2)) {
				out.println((i+1)+" "+(i+3));
				return;
			}
		}
		out.println("-1 -1");
	}

	public static void main(String[] args) {
		debug = args.length > 0;
		long start = System.nanoTime();

		solve();
		out.flush();

		long end = System.nanoTime();
		dump((end - start) / 1000000 + " ms");
		in.close();
		out.close();
	}

	static void dump(Object... o) { if (debug) System.err.println(Arrays.deepToString(o)); }
}