#include<bits/stdc++.h>
using namespace std;
const int N=1e6+5;
int vis[N],in[N];
int main(){
	int n,x,mx=0,ans=0;scanf("%d",&n);
	for(int i=0;i<n;i++)scanf("%d",&x),mx=max(mx,x),in[x]++;
	for(int i=1;i<=mx;i++){
		if(!in[i])continue;
		if(!vis[i]&&in[i]==1)ans++;
		for(int j=i;j<=mx;j+=i)vis[j]=1;
	}
	printf("%d",ans);
	return 0;
}
### Prompt

Develop a solution in CPP 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;
const int N=1e6+5;
int vis[N],in[N];
int main(){
	int n,x,mx=0,ans=0;scanf("%d",&n);
	for(int i=0;i<n;i++)scanf("%d",&x),mx=max(mx,x),in[x]++;
	for(int i=1;i<=mx;i++){
		if(!in[i])continue;
		if(!vis[i]&&in[i]==1)ans++;
		for(int j=i;j<=mx;j+=i)vis[j]=1;
	}
	printf("%d",ans);
	return 0;
}
```