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 solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}
void solve() {
  int n, x;
  cin >> n >> x;
  if (x == 10) {
    if (n == 1) {
      cout << -1;
    } else {
      for (int i = 0; i < n - 1; i++) {
        cout << 1;
      }
      cout << 0;
    }
  } else {
    for (int i = 0; i < n; i++) {
      cout << x;
    }
  }
}