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.util.Scanner;


public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		int digit,divis;
		Scanner scan=new Scanner(System.in);
		
		digit=scan.nextInt();
		divis=scan.nextInt();
		scan.close();
		
		if(digit==1 && divis==10){
			System.out.println(-1);
			System.exit(0);
		}
		switch(divis){
		case 2:
			for(int i=0;i<=digit-1;i++){
				System.out.print(2);
			}
			break;
		case 3:
			for(int i=0;i<=digit-1;i++){
				System.out.print(3);
			}
			break;
		case 4:
			for(int i=0;i<=digit-1;i++){
				System.out.print(4);
			}
			break;
		case 5:
			for(int i=0;i<=digit-1;i++){
				System.out.print(5);
			}
			break;
		case 6:
			for(int i=0;i<=digit-1;i++){
				System.out.print(6);
			}
			break;
		case 7:
			for(int i=0;i<=digit-1;i++){
				System.out.print(7);
			}
			break;
		case 8:
			for(int i=0;i<=digit-1;i++){
				System.out.print(8);
			}
			break;
		case 9:
			for(int i=0;i<=digit-1;i++){
				System.out.print(9);
			}
			break;
        case 10:
        	System.out.print(1);
        	for(int i=1;i<=digit-1;i++){
				System.out.print(0);
			}
			break;
		}
	}

}