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() {
  int n, t;
  cin >> n >> t;
  if (n == 1 && t == 10)
    cout << -1;
  else {
    switch (t) {
      case 2:
        for (int i = 0; i < n; i++) cout << 2;
        break;
      case 3:
        for (int i = 0; i < n; i++) cout << 3;
        break;
      case 4:
        for (int i = 0; i < n; i++) cout << 4;
        break;
      case 5:
        for (int i = 0; i < n; i++) cout << 5;
        break;
      case 6:
        for (int i = 0; i < n; i++) cout << 6;
        break;
      case 7:
        for (int i = 0; i < n; i++) cout << 7;
        break;
      case 8:
        for (int i = 0; i < n; i++) cout << 8;
        break;
      case 9:
        for (int i = 0; i < n; i++) cout << 9;
        break;
      case 10:
        for (int i = 0; i < n - 1; i++) cout << 1;
        cout << 0;
        break;
    };
  }
  return 0;
}