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 b, c, d, e, f, g, h, i, j, k, l, m, n, t;
  scanf("%d%d", &n, &t);
  int a[n];
  if (t == 10 && n == 1) {
    printf("-1\n");
    return 0;
  }
  if (t < 10) {
    for (i = 0; i < n; i++) {
      a[i] = t;
    }
    for (i = 0; i < n; i++) {
      printf("%d", a[i]);
    }
    printf("\n");
    return 0;
  }
  a[0] = 1;
  if (t == 10 && n > 1) {
    for (i = 1; i < n; i++) {
      a[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    printf("%d", a[i]);
  }
  printf("\n");
  return 0;
}