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 std::cerr;
using std::cin;
using std::cout;
using std::ios_base;
using std::log10;
using std::trunc;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, t;
  cin >> n >> t;
  if (n == 1 and t == 10) {
    cout << "-1"
         << "\n";
    return 0;
  }
  int digit = t % 10, ans[n];
  ans[n - 1] = digit;
  for (int i = n - 2; ~i; --i) ans[i] = digit == 0 ? 1 : digit;
  for (int i = 0; i < n; i++) cout << ans[i];
  cout << "\n";
  return 0;
}
