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;
long long power(long long b, long long exp, long long m) {
  long long ans = 1;
  b %= m;
  while (exp) {
    if (exp & 1) ans = (ans * b) % m;
    exp >>= 1;
    b = (b * b) % m;
  }
  return ans;
}
int main() {
  int i, n, t;
  cin >> n >> t;
  int ans = 0;
  if (t >= 2 && t <= 9) {
    for (i = 1; i <= n; i++) cout << t;
    cout << endl;
    return 0;
  } else if (t == 10) {
    if (n == 1) {
      cout << "-1" << endl;
      return 0;
    } else
      for (i = 1; i < n; i++) cout << "1";
    cout << "0" << endl;
  }
}