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 main() {
  int n, t;
  cin >> n >> t;
  if (n == 1) {
    if (t < 10)
      cout << t;
    else
      cout << -1;
    return 0;
  }
  vector<int> ans(n, 0);
  ans[0] = 1;
  int rest = 0;
  for (int x : ans) {
    rest = rest * 10 + x;
    rest %= t;
  }
  if (rest) ans.back() += t - rest;
  for (int x : ans) cout << x;
}