Problem: 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

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

    public class Main {
        public void solve() 
        {
            int n = nextInt();
            int t = nextInt();
            if (n == 1 && t == 10) System.out.print("-1");
            else
            if (t != 10)
            {
                for (int i = 0; i < n; i++)
                {
                    System.out.print(t);
                }
            } else {
                for (int i = 0; i < n - 1; i++)
                {
                    System.out.print("1");
                }
                System.out.print("0");
            }
        }
        public static void main(String[] args) {
            new Main().run();
        }

        BufferedReader br;
        StringTokenizer st;
        PrintWriter out;

        private void run() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
                //br = new BufferedReader(new FileReader("birthday.in"));
                //out = new PrintWriter(new FileWriter("birthday.out"));
                solve();
                br.close();
                out.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

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

        private String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                    return "END_OF_FILE";
                }
            }
            return st.nextToken();
        }
        int binary_search(int mass[], int key, int min, int max)
        {
            int midpoint = (min + max) / 2;
            if (min == max)
            {   
                if (min == 0)
                return -1;
                else 
                return midpoint;
            } else 
            if (mass[midpoint] == key)
            {
                return midpoint; 
            }
            else 
            if (key > mass[midpoint]) return binary_search(mass, key, midpoint, max);
            else 
            return binary_search(mass, key, min, midpoint);
        }
    }