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

#define N 1000100

int n, cnt[N];

int main() {
    scanf("%d", &n);
    for(int i = 1, a; i <= n; i ++) scanf("%d", &a), cnt[a] ++;
    int ans = 0;
    for(int i = 1; i < N; i ++) {
        if(!cnt[i]) continue;
        if(cnt[i] == 1) ans ++;
        for(int j = 2 * i; j < N; j += i) cnt[j] = 0;
    }
    printf("%d\n", ans);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 N 1000100

int n, cnt[N];

int main() {
    scanf("%d", &n);
    for(int i = 1, a; i <= n; i ++) scanf("%d", &a), cnt[a] ++;
    int ans = 0;
    for(int i = 1; i < N; i ++) {
        if(!cnt[i]) continue;
        if(cnt[i] == 1) ans ++;
        for(int j = 2 * i; j < N; j += i) cnt[j] = 0;
    }
    printf("%d\n", ans);
}

```