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;
char a[105][105];
int main() {
  long long int n, t, ans = 0, i;
  cin >> n >> t;
  if (n == 1) {
    if (t != 10)
      cout << t << endl;
    else
      cout << -1 << endl;
  } else {
    long long int ans1 = 2, ans2 = 9;
    if (t == 2 || t == 5 || t == 10 || t == 4 || t == 1) {
      cout << ans1;
      for (i = 1; i < n; i++) cout << 0;
    } else if (t == 9 || t == 3) {
      cout << ans2;
      for (i = 1; i < n; i++) cout << ans2;
    } else if (t == 8) {
      ans = 24;
      cout << ans;
      for (i = 2; i < n; i++) cout << 0;
    } else if (t == 7) {
      ans = 49;
      cout << ans;
      for (i = 2; i < n; i++) cout << 0;
    } else {
      ans = 36;
      cout << ans;
      for (i = 2; i < n; i++) cout << 0;
    }
  }
  return 0;
}