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;
void calculate() {
  string s;
  cin >> s;
  long long m = count(s.begin(), s.end(), '0');
  long long n = count(s.begin(), s.end(), '1');
  if ((min(m, n)) % 2 == 0) {
    cout << "NET" << endl;
  } else {
    cout << "DA" << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t, n;
  cin >> n >> t;
  if (n == 1 && t == 10) {
    cout << -1;
  } else if (t == 10) {
    cout << 1;
    for (long long i = 0; i < n - 1; i++) cout << 0;
  } else {
    for (long long i = 0; i < n; i++) cout << t;
  }
  cout << endl;
}