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 poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 1e9 + 7;
int main() {
  long long n, t;
  cin >> n >> t;
  if (t == 10) {
    if (n == 1)
      return cout << -1 << '\n', 0;
    else {
      cout << 1;
      n--;
      while (n--) cout << 0;
      cout << '\n';
      return 0;
    }
  }
  while (n--) cout << t;
  cout << '\n';
  return 0;
}