Olesya and Rodion (思维）

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

思路：如果给定的n就输出n个t这样就整除为n个1，但是有一种特殊的n为1，t为10，是无法找到的，特判一下

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
int main() {

	int n,k;
	cin>>n>>k;
	if(n==1&&k==10) {
		cout<<"-1"<<endl;
		return 0;
	}
	if(k>=2&&k<=9) {
		for(int t=0; t<n; t++) {
			cout<<k;
		}
	} else {
		for(int t=0; t<n; t++) {
			if(t==0) {
				cout<<"1";
			} else {
				cout<<"0";
			}
		}
	}

	return 0;
}