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

int i, j, n, ans;
int flag[1000001];

int main()
{
    scanf("%d", &n);
    int a[n];
    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    sort(a, a+n);
    for (i = 0; i < n; i++) {
        for (j = a[i]; j <= a[n-1]; j += a[i]) {
            flag[j]++;
        }
    }
    for (i = 0; i < n; i++) {
        if (flag[a[i]] == 1) ans++;
    }
    printf("%d\n", ans);
    return 0;
}
### 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;

int i, j, n, ans;
int flag[1000001];

int main()
{
    scanf("%d", &n);
    int a[n];
    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    sort(a, a+n);
    for (i = 0; i < n; i++) {
        for (j = a[i]; j <= a[n-1]; j += a[i]) {
            flag[j]++;
        }
    }
    for (i = 0; i < n; i++) {
        if (flag[a[i]] == 1) ans++;
    }
    printf("%d\n", ans);
    return 0;
}
```