Problem:
Score : 400 points

Problem Statement

Given is a number sequence A of length N.

Find the number of integers i \left(1 \leq i \leq N\right) with the following property:

For every integer j \left(1 \leq j \leq N\right) such that i \neq j , A_j does not divide A_i.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the answer.

Sample Input 1

5
24 11 8 3 16

Sample Output 1

3

The integers with the property are 2, 3, and 4.

Sample Input 2

4
5 5 5 5

Sample Output 2

0

Note that there can be multiple equal numbers.

Sample Input 3

10
33 18 45 28 8 19 89 86 2 4

Sample Output 3

5

Solution in C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdlib.h>

int compare_int(const void *a, const void *b)
{
  return *(int*)a - *(int*)b;
}

int main()
{
  int n;
  scanf("%d", &n);
  int a[n], i, j;
  for (i = 0; i < n; i++)
    scanf("%d", &a[i]);
  qsort(a, n, sizeof(int), compare_int);
  int cnt = 0;
  int flg;
  for (i = 0; i < n; i++)
  {
    flg = 0;
    if (a[i] == a[i + 1] || a[i] == a[i - 1])
      flg = 1;
    if (flg == 0)
    {
      for (j = 0; j < i; j++)
      {
        if (a[i] % a[j] == 0)
        {
          flg = 1;
          break;
        }
      }
    }
    if (flg == 0)
      cnt++;
  }
  printf("%d", cnt);
}