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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(0);
  cout << fixed;
  int n, t;
  bool flag = false;
  cin >> n >> t;
  for (int i = 2; i <= 10; i++) {
    if (n == 1 && t == 10) {
      continue;
    } else if (i % t == 0) {
      flag = true;
      if (t == 10) {
        for (int j = 0; j < n - 1; j++) {
          cout << 1;
        }
        cout << 0;
      } else {
        for (int j = 0; j < n; j++) {
          cout << t;
        }
      }
      break;
    }
  }
  if (!flag) {
    cout << -1;
  }
  return 0;
}