#include<bits/stdc++.h>
using namespace std;
using ll=long long;
int n,a[1<<18],ans,c[1<<20],b[1<<20];
basic_string<int> v[1<<20];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",a+i),c[a[i]]++;
	for(int i=1;i<(1<<20);i++)for(int j=i;j<(1<<20);j+=i)b[j]+=c[i];
	for(int i=1;i<=n;i++)if(b[a[i]]==1)ans++;
	printf("%d\n",ans);
	return 0;
}
### Prompt

Please create a solution in cpp 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<bits/stdc++.h>
using namespace std;
using ll=long long;
int n,a[1<<18],ans,c[1<<20],b[1<<20];
basic_string<int> v[1<<20];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",a+i),c[a[i]]++;
	for(int i=1;i<(1<<20);i++)for(int j=i;j<(1<<20);j+=i)b[j]+=c[i];
	for(int i=1;i<=n;i++)if(b[a[i]]==1)ans++;
	printf("%d\n",ans);
	return 0;
}
```