Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  if (m > 1 && m < 10) {
    int ar[n];
    ar[0] = m;
    for (int i = 1; i < n; i++) {
      ar[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      cout << ar[i];
    }
  } else {
    if (n > 1) {
      int ar1[n];
      for (int i = 0; i < n - 1; i++) {
        ar1[i] = 1;
      }
      ar1[n - 1] = 0;
      for (int i = 0; i < n; i++) {
        cout << ar1[i];
      }
    } else {
      cout << "-1";
    }
  }
  return 0;
}