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.*;
import java.util.*;

public class Codeforces {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        int n = in.nextInt();
        int t = in.nextInt();

        if (n == 1 && t == 10) {
            out.println(-1);
            out.close();
            return;
        }

        int first = t == 10 ? 1 : t;

        StringBuilder sb = new StringBuilder();
        sb.append(first);

        for (int i = 0; i < n - 1; i++) {
            sb.append(0);
        }

        out.println(sb.toString());
        out.close();
    }

    public static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int[] nextIntArr(int n) {
            int[] arr = new int[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j] = nextInt();
            }

            return arr;
        }

        public long[] nextLongArr(int n) {
            long[] arr = new long[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j] = nextLong();
            }

            return arr;
        }
    }
}