#include<bits/stdc++.h>

using namespace std;
int n,a[2000005],ans,ada[2000005];
int main(){
	ios_base::sync_with_stdio(NULL);
	cin.tie(0);cout.tie(0);
	cin>>n;
	memset(ada,0,sizeof ada);
	for(int i=1;i<=n;i++){
		cin>>a[i];
		ada[a[i]]++;
	}
	sort(a+1,a+n+1);
	ans=0;
	for(int i=1;i<=n;i++){
		if(ada[a[i]]==-1) continue;
		for(int j=2*a[i];j<=a[n];j+=a[i]){
			ada[j]=-1;
		}
		if(ada[a[i]]==1) ans++;
		ada[a[i]] = -1;
	}
	cout<<ans<<endl;
}

### 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;
int n,a[2000005],ans,ada[2000005];
int main(){
	ios_base::sync_with_stdio(NULL);
	cin.tie(0);cout.tie(0);
	cin>>n;
	memset(ada,0,sizeof ada);
	for(int i=1;i<=n;i++){
		cin>>a[i];
		ada[a[i]]++;
	}
	sort(a+1,a+n+1);
	ans=0;
	for(int i=1;i<=n;i++){
		if(ada[a[i]]==-1) continue;
		for(int j=2*a[i];j<=a[n];j+=a[i]){
			ada[j]=-1;
		}
		if(ada[a[i]]==1) ans++;
		ada[a[i]] = -1;
	}
	cout<<ans<<endl;
}

```