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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  unsigned int size, divs;
  cin >> size >> divs;
  vector<char> numbers(size);
  if (divs < 10) {
    for (int i = 0; i < size; i++) {
      numbers[i] = divs + 48;
    }
  } else {
    numbers[0] = 49;
    for (int i = 1; i < size; i++) {
      numbers[i] = 48;
    }
  }
  if (size < 2 && divs == 10) {
    cout << "-1" << endl;
    return 0;
  }
  for (int i = 0; i < size; i++) {
    cout << numbers[i];
  }
  cout << endl;
}