#include<bits/stdc++.h>
using namespace std;
int main(){
  int n;
  cin>>n;
  int a[n];
  map<int,int> m;
  for(int i=0;i<n;i++){
    cin>>a[i];
    m[a[i]]++;
  }
  sort(a,a+n);
  int max=a[n-1];
 	for(int i=0;i<n;i++){
      	for(int j=2;j*a[i]<=max;j++){
        	m[j*a[i]]=-1;
      	}
    }
  	int count=0;
  	for(int i=0;i<n;i++){
      if(m[a[i]]==1)count++;
    }
  	printf("%d",count);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

Examples

Input

5
24 11 8 3 16


Output

3


Input

4
5 5 5 5


Output

0


Input

10
33 18 45 28 8 19 89 86 2 4


Output

5

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
  int n;
  cin>>n;
  int a[n];
  map<int,int> m;
  for(int i=0;i<n;i++){
    cin>>a[i];
    m[a[i]]++;
  }
  sort(a,a+n);
  int max=a[n-1];
 	for(int i=0;i<n;i++){
      	for(int j=2;j*a[i]<=max;j++){
        	m[j*a[i]]=-1;
      	}
    }
  	int count=0;
  	for(int i=0;i<n;i++){
      if(m[a[i]]==1)count++;
    }
  	printf("%d",count);
  return 0;
}

```