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: Accepted)

#include<stdio.h>
//小さい順にソート
int c[200002];
void merge(int i,int j,int *s){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[d]=s[b];
        b++;
      }
      else if(j<b){
        c[d]=s[a];
        a++;
      }
      else if(s[a]<s[b]){
        c[d]=s[a];
        a++;
      }
      else{
        c[d]=s[b];
        b++;
      }
    }
    for(d=i;d<=j;d++)
      s[d]=c[d];
  }
}

int main(){
  int n,i;
  scanf("%d",&n);
  int a[n+1];
  for(i=0;i<n;i++)
    scanf("%d",&a[i]);
  merge(0,n-1,a);
  int dp[a[n-1]+1];
  for(i=1;i<=a[n-1];i++)
    dp[i]=0;
  for(i=0;i<n;i++){
    int j=2;
    if(a[i]==a[i-1] && i!=0)
      j=1;
    while(j*a[i]<=a[n-1]){
      dp[j*a[i]]=1;
      j++;
    }
  }
  int count=0;
  for(i=0;i<n;i++)
    if(dp[a[i]]==0)
      count++;
  printf("%d\n",count);
  return 0;
}