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>
const int INF = 1e9;
const long long lINF = 1e18;
const double EPS = 1e-12;
using namespace std;
int n, t;
int main() {
  cin >> n >> t;
  if (n == 1) {
    if (t == 10) {
      cout << -1 << endl;
    } else {
      cout << t << endl;
    }
  } else {
    int res = 1;
    for (int i = 0; i < n - 1; i++) {
      res = (res * 10) % t;
    }
    printf("1");
    for (int i = 0; i < n - 2; i++) {
      printf("0");
    }
    printf("%c\n", '0' + (char)((t - res) % t));
  }
  return 0;
}