#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main(){
	ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);
	vector<int> vec(1000000+1);
	int n,x; cin>>n;
	for(int i=1;i<=n;++i){
		cin>>x;
		++vec[x];
	}
	int cnt=0;
	for(int i=1;i<=1000000;++i){
		if(vec[i]==0) continue;
		if(vec[i]==1) ++cnt;
		for(int j=2*i;j<=1000000;j+=i){
			vec[j]=0;
		}
	}
	cout<<cnt<<"\n";
	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;
#define ll long long
int main(){
	ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);
	vector<int> vec(1000000+1);
	int n,x; cin>>n;
	for(int i=1;i<=n;++i){
		cin>>x;
		++vec[x];
	}
	int cnt=0;
	for(int i=1;i<=1000000;++i){
		if(vec[i]==0) continue;
		if(vec[i]==1) ++cnt;
		for(int j=2*i;j<=1000000;j+=i){
			vec[j]=0;
		}
	}
	cout<<cnt<<"\n";
	return 0;
}

```