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 numdigits(long long n) { return floor(log10(n)) + 1; }
int main() {
  int n;
  cin >> n;
  int t;
  cin >> t;
  long long num = round(pow(10, (n - 1)));
  bool found = false;
  if (num < 0) {
    if (t == 10) {
      cout << 1;
      for (int i = 1; i < n; i++) cout << 0;
    } else {
      for (int i = 0; i < n; i++) cout << t;
    }
    found = true;
  }
  while (!found) {
    if (numdigits(num) != n) {
      found = true;
      cout << -1;
    } else if (num % t != 0)
      num++;
    else {
      cout << num;
      found = true;
    }
  }
}
