#include<bits/stdc++.h>
using namespace std;
const int MX = 1000005;

int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(MX);
  for (int &i : a) cin >> i;
  a.push_back(0);
  sort(a.begin(), a.end());
  a.push_back(0);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i - 1] == a[i]) continue;
    for (int j = a[i]; j < MX - 2; j += a[i]) b[j]++;
    if (b[a[i]] == 1 and a[i] != a[i + 1]) ans++;
  }
  cout << ans << endl;
  return 0;
}
### 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;
const int MX = 1000005;

int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(MX);
  for (int &i : a) cin >> i;
  a.push_back(0);
  sort(a.begin(), a.end());
  a.push_back(0);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i - 1] == a[i]) continue;
    for (int j = a[i]; j < MX - 2; j += a[i]) b[j]++;
    if (b[a[i]] == 1 and a[i] != a[i + 1]) ans++;
  }
  cout << ans << endl;
  return 0;
}
```