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
#include <bits/stdc++.h>
using namespace std;
int c[100000];
void trans(long long n, int a[]) {
  long long num = n, i = 0, d;
  while (num) {
    d = num % 10;
    num /= 10;
    c[i] = d;
    i++;
  }
  int j = 0;
  for (int k = i - 1; k >= 0; k--) {
    a[k] = c[j];
    j++;
  }
}
int county(long long n) {
  int d, i = 0;
  while (n) {
    d = n % 10;
    n /= 10;
    i++;
  }
  return i;
}
int main() {
  long long n, t;
  int a[100000];
  cin >> n >> t;
  if (county(t) <= n) {
    trans(t, a);
    int c = n - county(t);
    for (int i = county(t); i <= c; i++) {
      a[i] = 0;
    }
    for (int i = 0; i < n; i++) cout << a[i];
  } else
    cout << "-1";
}
