Codeforces Round #324 (Div. 2) A

A. Olesya and Rodion

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

题意； 输出一个数  长度为n并且为t的倍数  不存在 则输出-1

题解： n个t  组成的数一定是t的倍数  
      考虑特殊  t=10的情况  （水）

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<cmath>
#define ll __int64 
#define pi acos(-1.0)
#define mod 1000000007
using namespace std;
int n,t;
int main()
{
    scanf("%d %d",&n,&t);
    if(t<10)
    {
        for(int i=1;i<=n;i++)
            printf("%d",t);
            cout<<endl;
    }
    if(t==10)
    {
        if(n==1)
        cout<<"-1"<<endl;
        else
        {
            for(int i=1;i<n;i++)
            printf("1",t);    
             cout<<"0"<<endl;
        }
    }
    return 0;
 }