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 number(int n) {
  int cnt = 0;
  while (n > 0) {
    if (n % 10 == 0) {
      cnt++;
      n = n / 10;
    } else {
      n = n / 10;
      cnt++;
    }
  }
  return cnt;
}
int main() {
  int n;
  cin >> n;
  int t;
  cin >> t;
  if (n == 1) {
    if (t < 10) {
      cout << t;
      return 0;
    } else {
      cout << "-1";
      return 0;
    }
  } else if (n == number(t)) {
    cout << t;
  } else {
    cout << t;
    for (int i = 0; i < n - number(t); i++) {
      cout << "0";
    }
  }
}