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.
OutputPrint 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
AC代码(一开始想的是后面补零后来发现直接输出那个数就好了因为题目范围太友好)
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, t, i;
while(scanf("%d %d",&n, &t)!=EOF)
{
if(t==10)
{
if(n==1)
printf("-1\n");
else
{
printf("%d",t);
for(i = 0; i<n-2; i++)
{
printf("0");
}
printf("\n");
}
}
else
{
printf("%d",t);
for(i = 0; i<n-1; i++)
printf("%d",t);
printf("\n");
}
}
return 0;
}