#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n; scanf("%d",&n);
	vector<int> a(n);
	rep(i,n) scanf("%d",&a[i]);

	vector<int> f(1000001);
	rep(i,n) f[a[i]]++;
	for(int x=1;x<=1000000;x++) if(f[x]>0) {
		for(int y=2*x;y<=1000000;y+=x) f[y]=0;
	}
	printf("%ld\n",count(f.begin(),f.end(),1));

	return 0;
}

### Prompt

Your task is to create 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 <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n; scanf("%d",&n);
	vector<int> a(n);
	rep(i,n) scanf("%d",&a[i]);

	vector<int> f(1000001);
	rep(i,n) f[a[i]]++;
	for(int x=1;x<=1000000;x++) if(f[x]>0) {
		for(int y=2*x;y<=1000000;y+=x) f[y]=0;
	}
	printf("%ld\n",count(f.begin(),f.end(),1));

	return 0;
}

```