Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

5
24 11 8 3 16

Sample Output 1

3

The integers with the property are 2, 3, and 4.

Sample Input 2

4
5 5 5 5

Sample Output 2

0

Note that there can be multiple equal numbers.

Sample Input 3

10
33 18 45 28 8 19 89 86 2 4

Sample Output 3

5

Solution in C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include <assert.h>
#include <stdio.h>
#include <string.h>

/*
	入力例：
	3
	1 1 3
*/
int main(){
	int iDataNum,st,sx,sy,i;
	int ct,cx,cy;
	int strInput[1];
	for(int i=0;i<1;i++){
	 scanf("%d", &strInput[i]);
	}

	int strInput2[strInput[0]];
	for(int i=0;i<strInput[0];i++){
	 scanf("%d", &strInput2[i]);
	}

	int cnt[1000001]={0};
	int iRetCnt;
	iRetCnt=0;
	for(int i=0;i<strInput[0];i++){

		for(int j=strInput2[i];j<1000001;j+=strInput2[i]){
			if(cnt[j]>=1){
				cnt[j]=2;
				continue;
			}
			cnt[j]=1;
		}
	}
	for(int i=0;i<strInput[0];i++){
		if(cnt[strInput2[i]]==1){
			iRetCnt++;
		}
	}
	printf("%d\n",iRetCnt);

  return 0;
}