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 n, t, i = 1;
string x, res;
string toStr(int x) {
  string ans = "";
  while (x) {
    ans += ((x % 10) + '0');
    x /= 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int main() {
  cin >> n >> t;
  x = toStr(t);
  if (x.length() > n) {
    cout << -1;
    return 0;
  }
  res = x;
  for (int i = 0; i < n - x.length(); i++) {
    res += '0';
  }
  cout << res;
}