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
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  int digits = 0;
  int d = k;
  while (d > 0) {
    digits++;
    d /= 10;
  }
  if (digits <= n) {
    int a[100] = {0};
    a[0] = 1;
    int i = digits - 1, t = k;
    while (k > 0) {
      a[i--] = k % 10;
      k /= 10;
    }
    for (i = 0; i < n; i++) cout << a[i];
  } else
    cout << "-1";
  return 0;
}
