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 Solution implements Runnable {
    private void solve() throws IOException {
        //String s = nextToken();
        int n = nextInt();
        int t = nextInt();

        if (n < String.valueOf(t).length()) {
            writer.println(-1);
            return;
        }
        
        String ans = "";
        for (int i = 1; i < n; i++) {
            ans += "0";
        }
        
        ans = t + ans;
        while (ans.length() > n)
            ans = ans.substring(0,ans.length()-1);
        
       writer.println(ans);
        
       
       
             
    }


    public static void main(String[] args) {
        new Solution().run();
    }

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            writer = new PrintWriter(System.out);
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }
}