#include "bits/stdc++.h"
using namespace std;

const int N=2e5+20,MAX=1e6+20;

int n,x,f[MAX];
bool ok[MAX];

int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&x),f[x]++;
	for(int i=1;i<MAX;i++) ok[i]=true;

	for(int i=1;i<MAX;i++)
		if(f[i])
			for(int j=2*i;j<MAX;j+=i)
				ok[j]=false;
	int ans=0;
	for(int i=1;i<MAX;i++)
		if(ok[i] and f[i]==1) ans++;
	printf("%d",ans);
}
### Prompt

Your task is to create a CPP solution to the following problem:

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;

const int N=2e5+20,MAX=1e6+20;

int n,x,f[MAX];
bool ok[MAX];

int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&x),f[x]++;
	for(int i=1;i<MAX;i++) ok[i]=true;

	for(int i=1;i<MAX;i++)
		if(f[i])
			for(int j=2*i;j<MAX;j+=i)
				ok[j]=false;
	int ans=0;
	for(int i=1;i<MAX;i++)
		if(ok[i] and f[i]==1) ans++;
	printf("%d",ans);
}
```