#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, a[200005], out[1000005], ans;
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%d", a+i);
		if(out[a[i]] == 0){
			for(int t = a[i]; t <= 1000000; t += a[i]) out[t]++;
		}else out[a[i]]++;
	}
	rep(i,n) if(out[a[i]] == 1) ans++;
	printf("%d\n", ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 rep(i,n) for(int i=0; i<(n); i++)
int n, a[200005], out[1000005], ans;
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%d", a+i);
		if(out[a[i]] == 0){
			for(int t = a[i]; t <= 1000000; t += a[i]) out[t]++;
		}else out[a[i]]++;
	}
	rep(i,n) if(out[a[i]] == 1) ans++;
	printf("%d\n", ans);
}
```