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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

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

int main()
{
   int n;
   scanf("%d", &n);
   int a[n];
   int marcador[1000000] = {0};
   int result = 0, valor, igual = 0;

   for(int i=0; i<n; i++)
   {
      scanf(" %d", &a[i]);
   }
   qsort(a,n,sizeof(int),comparador);

   for(int i=0; i<n; i++)
   {
      if(marcador[a[i]] == 0)
      {
         result++;
         for(int j=2;; j++)
         {
            valor = a[i]*j;
            if(valor > 1000000)
               break;
            marcador[valor] = 1;
         }
      }
      if(a[i] == a[0])
         igual++;
   }
   if(igual == n)
   {
      printf("%d\n", 0);
      return 0;
   }

   printf("%d\n", result);
   return 0;
}