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
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Scanner;

public class A {

    public static void main(String[] args) {
        final Scanner sc = new Scanner(new BufferedInputStream(System.in));
        /*
            100 digits => 10^100
            2^10 => 10^3
            10^100 ~ (10^3)^33 ~ ((2^10)^3)^33 ~ 2^(30 * 33) ~ 2^99
         */
        int N = sc.nextInt();
        String T = sc.next();
        BigInteger res = new BigInteger(T);
        BigInteger mult = new BigInteger(T);

        BigInteger limit = new BigInteger("10").pow(N);
        BigInteger lower = new BigInteger("10").pow(N-1);
        while(res.multiply(mult).compareTo(limit) == -1)
            res = res.multiply(mult);

        System.out.println((res.compareTo(lower) != -1 && res.compareTo(limit) == -1) ? res.toString() : -1);
    }
}