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;
int main() {
  int n, m;
  cin >> n >> m;
  long long k = pow(10, n - 1);
  if (n > 9) {
    if (m == 10) {
      m = 1;
      n--;
      for (int i = 0; i < n; i++) {
        cout << m;
      }
      cout << 0;
      return 0;
    }
    for (int i = 0; i < n; i++) {
      cout << m;
    }
    return 0;
  }
  if (k < 0) {
    cout << pow(10, n - 1) * m;
    return 0;
  }
  if (m == 10 && n != 1) {
    cout << k;
    return 0;
  }
  if (n != 1) {
    cout << k * m;
    return 0;
  }
  if (n == 1 && m < 10) {
    cout << m;
    return 0;
  } else {
    cout << "-1";
    return 0;
  }
}