#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[200005], dp[1000005];
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]), dp[a[i]]++;
    for(int i=1000000;i>=1;i--){
        for(int j=i+i;j<=1000000;j+=i) dp[j] += dp[i];
    }
    int ans = 0;
    for(int i=1;i<=n;i++){
        if(dp[a[i]] == 1) ans++;
    }
    printf("%d\n", ans);
}
### Prompt

Your challenge is to write 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 a[200005], dp[1000005];
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]), dp[a[i]]++;
    for(int i=1000000;i>=1;i--){
        for(int j=i+i;j<=1000000;j+=i) dp[j] += dp[i];
    }
    int ans = 0;
    for(int i=1;i<=n;i++){
        if(dp[a[i]] == 1) ans++;
    }
    printf("%d\n", ans);
}
```