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;
struct point {
  int x;
  int y;
};
int distance(point a, point b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
bool is_between(point a, point c, point b) {
  return distance(a, c) + distance(c, b) == distance(a, b);
}
int main() {
  int n, t;
  cin >> n >> t;
  if (n == 1 && t == 10)
    cout << "-1" << endl;
  else {
    if (t == 10) {
      cout << t;
      for (int i = 0; i < n - 2; i++) cout << "0";
    } else {
      cout << t;
      for (int i = 0; i < n - 1; i++) cout << "0";
    }
  }
  cout << endl;
  return 0;
}
