#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main(){
  int n;
  cin >> n ;
  vector<ll> a(n);
  for(int i= 0; i< n; i++)cin >> a[i];
  sort(a.begin(), a.end());
  ll amax = a[n-1];
  vector<ll> p(amax+5);

 for(int i =0; i< n; i++){
   for(int j = a[i]; j < amax + 5 ; j += a[i]){
      p[j]++;
    }
 }
ll ans = 0;
 for(auto x :a){
    if(p[x]==1) ans ++;
 }
cout << ans << endl;
  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 main(){
  int n;
  cin >> n ;
  vector<ll> a(n);
  for(int i= 0; i< n; i++)cin >> a[i];
  sort(a.begin(), a.end());
  ll amax = a[n-1];
  vector<ll> p(amax+5);

 for(int i =0; i< n; i++){
   for(int j = a[i]; j < amax + 5 ; j += a[i]){
      p[j]++;
    }
 }
ll ans = 0;
 for(auto x :a){
    if(p[x]==1) ans ++;
 }
cout << ans << endl;
  return 0;
}
```