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 int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int n, t;
  cin >> n >> t;
  if (n == 1 && t >= 10) return cout << -1, 0;
  if (t == 10) {
    putchar('1');
    for (int i = 1; i < n; i++) putchar('0');
  } else
    for (int i = 0; i < n; i++) putchar(t + '0');
  return 0;
}