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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) throws IOException {

		BufferedReader reader = new BufferedReader(new InputStreamReader(
				System.in));
		StringTokenizer stringTokenizer = new StringTokenizer(reader.readLine());
		int n = Integer.parseInt(stringTokenizer.nextToken()), d = Integer
				.parseInt(stringTokenizer.nextToken());
		String ans = n * d + "";
		StringBuilder builder = new StringBuilder();
		builder.append(ans);
		for (int i = 0; i < n - ans.length(); i++) {
			builder.insert(0, "1");
		}
		BigInteger bigN = new BigInteger(builder.toString());
		BigInteger bigD = new BigInteger(d + "");
		// System.out.println(builder);
		while (!(bigN.mod(bigD).equals(BigInteger.ZERO)) && d < 100) {
			d++;
			bigN = bigN.add(BigInteger.ONE);
			// System.out.println(bigN);
		}
		if ((bigN.mod(bigD)).equals(BigInteger.ZERO)
				&& bigN.toString().length() <= n) {
			System.out.println(bigN);
		} else
			System.out.println(-1);
	}
}
