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, t, ans;
  cin >> n >> t;
  if (n == 1) {
    if (t < 10)
      cout << t;
    else
      cout << -1;
  } else {
    if (t == 2 || t == 4 || t == 8) {
      if (n == 2)
        cout << "16";
      else {
        if (n == 3 && t == 8)
          cout << "104";
        else {
          cout << '1';
          for (int i = 0; i < n - 2; i++) cout << '0';
          char u = '0' + t;
          cout << u;
        }
      }
    } else if (t == 3 || t == 6 || t == 9) {
      cout << '1';
      for (int i = 0; i < n - 2; i++) cout << '0';
      cout << '8';
    } else if (t == 5 || t == 10) {
      cout << "1";
      for (int i = 0; i < n - 1; i++) cout << '0';
    } else {
      char u;
      switch (n % 6) {
        case 0:
          u = '2';
          break;
        case 1:
          u = '6';
          break;
        case 2:
          u = '4';
          break;
        case 3:
          u = '5';
          break;
        case 4:
          u = '1';
          break;
        case 5:
          u = '3';
          break;
      }
      cout << '1';
      for (int i = 0; i < n - 2; i++) cout << '0';
      cout << u;
    }
  }
  return 0;
}
