Problem: 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

Solution (in JAVA):

Solution:
/*
* @Author Silviase(@silviasetitech)
* For ProCon
*/

import java.util.*;
import java.lang.*;
import java.math.*;

class Main{

        // usual use input int
        static int n;
        static int x;
        static int c;
        static int k;
        static int m;
        static int height;
        static int width;
        static int left;
        static int right;
        static int ans;
        //
        static String s;
        static String anss;
        // 
        static String[] sAr;
        //
        static long ansl;
        //
        static int[] intAr;
        //

        // for dfs
        static int[] w;
        static int[] ww;
        static boolean[] isvisited;
        //
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
             
        s = sc.next();
        boolean bool = false;
        int piv2 = 0;
        int piv3 = 0;
        for (int i = 0; i < s.length()-1; i++) {
            if (s.charAt(i) == s.charAt(i+1)){
                piv2 = i;
                bool = true;
                break;
            }else{
                continue;
            }
        }
        if (bool) {
            System.out.println( (piv2+1) + " " + (piv2+2) );
            return;
        }

        if (bool == false) {
            for (int i = 0; i < s.length()-2; i++) {
                if (s.charAt(i) == s.charAt(i+2)) {
                    piv3 = i;
                    bool = true;
                    break;
                }
            }
        }
        if (bool) {
            System.out.println( (piv3+1) +" " + (piv3+3) );
            return;
        }

        if (bool == false) {
            System.out.println(-1 + " " + -1);
            return;
        }

        
        sc.close();
    }


        public static int gcd_int(int a, int b) {
        
            if (a < b) {
                int tmpp = a;
                a = b;
                b = tmpp;
            }
            if (b == 0) {
                return a;
            }else{
                return gcd_int(b, a%b);
            }
        }

        public static long gcd_long(long a, long b) {
        
            if (a < b) {
                long tmpp = a;
                a = b;
                b = tmpp;
            }
            if (b == 0) {
                return a;
            }else{
                return gcd_long(b, a%b);
            }
        }

        public static int lcm_int(int a, int b) {
            int gcd = gcd_int(a,b);
            return a/gcd*b;
        }

        public static long lcm_long(long a, long b) {
            long gcd = gcd_long(a,b);
            return a/gcd*b;
        }

        public static void dfs(int placenow) {
            // if went all -> success! 
            // if not went all -> fail...
            
            /*
            dfs    
            Go All Place that there is way to and not having been yet. 

             if island 1 is start point, dfs(1);
            if there is way to island 2 and island 3,
            - island 2 changes non-visited -> visited, and dfs(2).
            - island 3 changes non-visited -> visited, and dfs(3).
    
            */

            isvisited[placenow] = true;
            boolean success = true;
            for (int i = 0; i < n; i++) {
                if (isvisited[i] == false) { // not go all place
                    success = false;
                    break;
                }
            }
            if (success) {
                ans++;
                isvisited[placenow] = false;
                return;
            }

            for (int i = 0; i < n; i++) {
                if (w[i] == placenow && isvisited[ww[i]] == false ) {
                    dfs(ww[i]);
                }else if(ww[i] == placenow && isvisited[w[i]] == false){
                    dfs(w[i]);
                }else{
                    continue;
                }
            }
            isvisited[placenow] = false;
            return;
        }

}