Problem: Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.

Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print  - 1.

Input

The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.

Output

Print one such positive number without leading zeroes, — the answer to the problem, or  - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

Examples

Input

3 2


Output

712

Solution (in JAVA):

Solution:
import java.util.Collections;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;

public class warmup{
	public static void main(String[] args) {
		Reader rd = new Reader();
		int n = rd.nextInt();
		int t = rd.nextInt();
		String str = "";
		if(n == 1 && t == 10){
			System.out.println(-1);
		}
		else{
			if(t == 10){
				str += "1";
			}
			else{
				str += Integer.toString(t);
			}
			while(str.length()< n){
				str += "0";
			}
		}
		System.out.println(str);
	}

	static void sort(int[] ar){
		ArrayList<Integer> ls = new ArrayList<>();
		for(int i=0; i<ar.length; i++) ls.add(ar[i]);
		Collections.sort(ls);
		for(int i =0; i<ls.size(); i++) ar[i] = ls.get(i);
	}
}

class Reader { 
	static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer tok = new StringTokenizer("");
	static String next() {
		while(!tok.hasMoreTokens()){
			try{
				tok = new StringTokenizer(bf.readLine());
			}
			catch(IOException e) {
				e.printStackTrace();
			}
		}
		return tok.nextToken();
	} 
	static int nextInt() {
		return Integer.parseInt(next());
	}
}