#include<bits/stdc++.h>
using namespace std;
const int N=1000005;
int a[N],v[N],n,ans;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)
	{
		int x=a[i];
		if(!v[x])
		{
			for(int j=x;j<=1000000;j+=x)
				v[j]=1;
			if(a[i]!=a[i+1])
				ans++;
		}
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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=1000005;
int a[N],v[N],n,ans;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)
	{
		int x=a[i];
		if(!v[x])
		{
			for(int j=x;j<=1000000;j+=x)
				v[j]=1;
			if(a[i]!=a[i+1])
				ans++;
		}
	}
	printf("%d\n",ans);
	return 0;
}
```