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.util.Scanner;

/**
 * Created by frillyfrufru on 22/12/15.
 * @codeforces.id A584
 * @codeforces.url http://codeforces.com/problemset/problem/584/A
 * @codeforces.keywords math
 * @codeforces.title Olesya and Rodion
 *
 * Given positive integers n and t, find a positive integer
 * consisting of n digits that is divisible by t, if such a
 * number exists.
 *
 * Input:
 * A single line with 1 <= n <= 100, 2 <= t <= 10.
 *
 * Output:
 * The answer to the problem or -1 if such a number does not exist.
 * If there are multiple answers, print one (any) of them.
 *
 * Discussion (divisibility rules)
 * ===============================
 * FYI:
 *
 * divisible by    rule
 *      2           the last digit is even
 *      3           the sum of the digits is divisible by 3
 *      4           the last two digits are divisible by 4
 *      5           the last digit is 0 or 5
 *      6           the number is divisible by 2 *and* 3
 *      7           double the last digit and subtract it from
 *                      the number. Then the number is divisible
 *                      by 7 if the answer is 0 or divisible by 7.
 *      8           the last three digits are divisible by 8
 *      9           the sum of the digits is divisible by 9
 *     10           the last digit is a 0
 *
 * For 2 through 9, just repeat t as a character n times.
 * The result abides by the divisibility rules above. For
 * t = 10, just concatenate "1" and (n-1) "0"s. Unless n
 * is 1, in which case the answer is -1. This is the only
 * time the answer is -1.
 */
public final class A584 {
    /** Do not instantiate. */
    private A584() { }
    
    public static void main(final String[] notUsed) {
        Scanner sin = new Scanner(System.in);
        int n = sin.nextInt();
        int t = sin.nextInt();

        String result = "";
        if (t < 10) {
            result = repeat(n, (char) (t + '0'));
        } else if (n > 1) {
            result = "1" + repeat(n - 1, '0');
        } else {
            result = "-1";
        }
        System.out.println(result);
    }

    private static String repeat(final int n, final char c) {
        char[] r = new char[n];
        for (int i = 0; i < n; i++) {
            r[i] = c;
        }
        return String.valueOf(r);
    }
}
