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 <math.h>
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define PI acos(-1)
typedef unsigned long long int ull;
typedef long long int ll;
typedef long double ld;

int main(void) {
  /* 標準入力フラグ */
  int   scan;

  /* 数入力 */
  int    answer;
  int    n;
  int    *a;
  answer = 0;
  scan = scanf("%d", &n);
  a = (int *)malloc(sizeof(int) * n);
  for (int i = 0; i < n; i++) {
    scan = scanf("%d", &a[i]);
  }

  int   *result;
  result = (int *)malloc(sizeof(int) * n);
  /* プログラムを書く */
  for (int i = 0; i < n; i++) {
    result[i] = 1;
    for (int j = 0; j < n; j++) {
      if (i != j && a[i] % a[j] == 0) { 
        result[i] = 0;
      }
    }
  }


  for (int i = 0; i < n; i++) {
    answer += result[i];
  }
  printf("%d", answer);

  return (0);
}