#include <cstdio>
#include <algorithm>
using namespace std;
int vis[1000010],a[200010],in[1000010];
int main(){
	int n,i,j,cnt,mn=1000010,mx=0,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]),vis[a[i]]++,in[a[i]]=1,mn=min(mn,a[i]),mx=max(mx,a[i]);
	for(i=1;i<=n;i++){
		cnt=1;
		while(1){
			cnt++;
			if(a[i]*cnt>mx)break;
			vis[a[i]*cnt]++;
		}
	}
	for(i=mn;i<=mx;i++)
		if(in[i]==1&&vis[i]==1)ans++;
	printf("%d\n",ans);
	return 0;
}
### Prompt

Generate 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 <cstdio>
#include <algorithm>
using namespace std;
int vis[1000010],a[200010],in[1000010];
int main(){
	int n,i,j,cnt,mn=1000010,mx=0,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]),vis[a[i]]++,in[a[i]]=1,mn=min(mn,a[i]),mx=max(mx,a[i]);
	for(i=1;i<=n;i++){
		cnt=1;
		while(1){
			cnt++;
			if(a[i]*cnt>mx)break;
			vis[a[i]*cnt]++;
		}
	}
	for(i=mn;i<=mx;i++)
		if(in[i]==1&&vis[i]==1)ans++;
	printf("%d\n",ans);
	return 0;
}
```