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 modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long mul_inv(long long a, long long b) { return modpow(a, b - 2, b); }
int main() {
  int t;
  int n, i;
  cin >> n >> t;
  string s = "";
  for (i = 0; i < n; i++)
    if (t != 10) {
      s += (char)('0' + t);
    } else
      s += '1';
  if (t == 10) s[n - 1] = '0';
  if (t == 10 && n == 1) {
    cout << "-1" << endl;
    return 0;
  }
  cout << s << endl;
  return 0;
}