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;
}