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, t, i;
  string s = "";
  scanf("%d%d", &n, &t);
  if (t == 10) {
    if (n == 1)
      printf("-1\n");
    else {
      for (i = 0; i < n - 1; i++) s += '1';
      s += '0';
      cout << s << "\n";
    }
  } else {
    for (i = 0; i < n; i++) {
      s += (char)('0' + t);
    }
    cout << s << "\n";
  }
  return 0;
}