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, m, count = 0;
  cin >> n >> m;
  string a;
  while (m) {
    count++;
    a += (m % 10) + 48;
    m /= 10;
  }
  reverse(a.begin(), a.end());
  if (n < count) {
    cout << "-1" << endl;
    return 0;
  }
  n -= count;
  while (n--) {
    a += "0";
  }
  cout << a << endl;
  return 0;
}
