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 olesya
{
public static void main(String args[]) throws IOException
{Scanner sc= new Scanner(System.in);
int n=Integer.parseInt(sc.next());
int t=Integer.parseInt(sc.next());
int in=0,h=0,temp=0,d=0,count=0;
int arr[]=new int[n];
arr[0]=1;
for(int c=1;c<n-1;c++)
arr[c]=0;

arr[n-1]=1;

int flag=0;


do{


	if(n==1&&t==10)
	break;
in=arr[0];

for(int c=0;c<n;c++)
{

	if(in%t!=0)
	d=1;

if(c!=n-1)
	{if(d==1)
in=10*(in%t)+arr[c+1];
else
in=arr[c+1];}
else
continue;


	d=0;
	}
	if(in%t==0)
	{flag=1;break;}
	else{
		if(arr[n-1]!=9)
	arr[n-1]++;
	else
	{arr[n-2]++;
	arr[n-1]=0;}

	}


if(count==100)
break;
}while(true);

if(flag==1)
for(int c=0;c<n;c++)
System.out.print(arr[c]);
else
System.out.println("-1");


}}