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;
void print(list<int> *gr, int n) {
  for (int i = 1; i < n; ++i) {
    cout << i << "->";
    for (auto nbr : gr[i]) {
      cout << nbr << " ";
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, t;
  cin >> n >> t;
  if (t == 10) {
    if (n == 1) {
      cout << -1 << endl;
      return 0;
    } else {
      cout << 1;
      while (--n) {
        cout << 0;
      }
      return 0;
    }
  }
  while (n--) {
    cout << t;
  }
}