#include <bits/stdc++.h>

using namespace std;

int mark[2000001], f[2000001];
int a[200001];

int cnt, n;

const int N = (int )1e6;
int main( ) {
	scanf("%d", &n);
	for (int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
		mark[a[i]] ++;
	}
	for (int i = 1; i <= N; i ++)
		for (int j = i; j <= N; j += i)
			f[j] += mark[i];
	
	for (int i = 1; i <= n; i ++)
		if (f[a[i]] == 1) ++ cnt;
	printf("%d\n", cnt);
	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;

int mark[2000001], f[2000001];
int a[200001];

int cnt, n;

const int N = (int )1e6;
int main( ) {
	scanf("%d", &n);
	for (int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
		mark[a[i]] ++;
	}
	for (int i = 1; i <= N; i ++)
		for (int j = i; j <= N; j += i)
			f[j] += mark[i];
	
	for (int i = 1; i <= n; i ++)
		if (f[a[i]] == 1) ++ cnt;
	printf("%d\n", cnt);
	return 0;
}

```