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;
const int MN = 100 * 100 * 100 * 10;
int n, t;
string ans;
int main() {
  cin >> n >> t;
  int mod = 0;
  for (int i = 0; i < n - 1; i++) {
    ans += '1';
    mod = mod * 10 + 1;
    mod %= t;
  }
  mod = mod * 10;
  mod %= t;
  if (mod == 0)
    ans += '0';
  else
    ans += '0' + t - mod;
  if (ans[0] == '0') ans[0] += t;
  if (n == 1 && t == 10)
    cout << -1;
  else
    cout << ans;
  return 0;
}