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

int main() {
    ll n,ans=0;
    cin>>n;
    vector<ll> a(n),num(1000100),div(1000100);
    for(int i=0; i<n; i++) {
        cin>>a[i];
        num[a[i]]++;
    }
    for(int i=1; i<=1000000; i++) {
        for(int j=i; j<=1000000; j+=i) {
            div[j] += num[i];
        }
    }
    for(int i=0; i<n; i++) {
        if(div[a[i]] == 1) ans++;
    }
    cout<< ans <<"\n";
}
### Prompt

Generate 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>
using namespace std;
typedef long long ll;

int main() {
    ll n,ans=0;
    cin>>n;
    vector<ll> a(n),num(1000100),div(1000100);
    for(int i=0; i<n; i++) {
        cin>>a[i];
        num[a[i]]++;
    }
    for(int i=1; i<=1000000; i++) {
        for(int j=i; j<=1000000; j+=i) {
            div[j] += num[i];
        }
    }
    for(int i=0; i<n; i++) {
        if(div[a[i]] == 1) ans++;
    }
    cout<< ans <<"\n";
}
```