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:
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
	public static void main(String[] args) { 
		/*
		 *
		 * */
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		String S = I.next();
		int N = S.length();
		int A = -2;
		int B = -2;
		for (int i = 1; i < N; i++) {
			if (S.charAt(i)==S.charAt(i-1)) {
				A=i-1;
				B=i;
			}
		}
		for (int i = 0; i+2<N; i++) {
			if (S.charAt(i)==S.charAt(i+2)) {
				A=i;
				B=i+2;
			}
		}
		A++;
		B++; //One based indexing now
		O.pln(A+" "+B);
	}	
	public static double max(double a, double b) {return Math.max(a, b);}
	public static double min(double a, double b) {return Math.min(a, b);}
	public static long min(long a, long b) {return Math.min(a,b);}
	public static long max(long a, long b) {return Math.max(a,b);}
	public static int min(int a, int b) {return Math.min(a,b);}
	public static int max(int a, int b) {return Math.max(a,b);}
	public static long abs(long x) {return Math.abs(x);}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return max(a,b);
		return GCD(min(a,b),max(a,b)%min(a,b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	public static int pop(long x) { //Returns number of bits within a number
		int cnt = 0;
		while (x>0) {
			if (x%2==1) cnt++;
			x/=2;
		}
		return cnt;
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
		double nextDouble() {return Double.parseDouble(next());}
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(double x) {w.println(x);w.flush();}
		void pln(boolean x) {w.println(x);w.flush();}
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void pln(BigInteger x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
		void p(BigInteger x) {w.print(x);w.flush();}
		void p(double x) {w.print(x);w.flush();}
		void p(boolean x) {w.print(x);w.flush();}
	}
}