codeforces 584A Olesya and Rodion

A. Olesya and Rodion

 

 

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.

Sample test(s)

Input

3 2

Output

712

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int a[11][105];
int main()
{
    int n,t;
    for(int i=2;i<=10;i++)
    {
        a[i][1]=1;
        for(int j=2;j<=100;j++)
            a[i][j]=(a[i][j-1]*10+1)%i;
    }
    scanf("%d%d",&n,&t);
    int x=a[t][n-1]*10;
    x%=t;
    for(int i=0;i<n-1;i++)printf("1");
    printf("%d\n",t-x);
    return 0;
}