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;
long long power(long long b, long long x) {
  long long sum = 1;
  for (int i = 0; i < x; i++) {
    sum *= b;
  }
  return sum;
}
int main() {
  int n;
  string t;
  cin >> n >> t;
  if (n == 1 && t == "10") {
    cout << -1 << endl;
    return 0;
  }
  string s = "";
  if (t == "10") {
    s += '1';
    for (int i = 1; i < n; i++) {
      s += '0';
    }
  } else {
    for (int i = 0; i < n; i++) {
      s += t;
    }
  }
  cout << s << endl;
}