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.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        char[] str = scan.next().toCharArray();
        int n = str.length;
        TreeMap<Double, Integer> sumIdx = new TreeMap<>();
        for (char c = 'a'; c <= 'z'; c++){
            double sum = 0;
            sumIdx.clear();
            sumIdx.put(0.0, -1);
            for (int i=0;i<n;i++) {
                sum += str[i] == c? 0.5: -0.5;
                Map.Entry<Double, Integer> lower = sumIdx.lowerEntry(sum);
                if (lower != null) {
                    if (lower.getValue() == i - 1) lower = sumIdx.lowerEntry(lower.getKey());
                    if (lower != null) {
                        System.out.println(lower.getValue() + 2 + " " + (i+1));
                        return;
                    }
                }
                sumIdx.putIfAbsent(sum, i);
            }
        }
        System.out.println("-1 -1");
    }
}
