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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program sample
    implicit none
    integer(8) ::c,i,j,m,n,k,a1,b1,c1
    integer(8)::x,b(1000000)
    integer(8),allocatable::a(:)
    read(*,*) n
    
    allocate(a(n))
    
    read(*,*)a
    do i=1,10**6
        b(i)=1
    end do
    do i=1,n
        x=a(i)
        do j=2,n/x
            b(x*j)=0
        end do
    end do
    m=0
    do i=1,n
        if (b(a(i))==1)then
            m=m+1
            b(a(i))=2
        else if (b(a(i))==2)then
            m=m-1
            b(a(i))=0
        end if
    end do

    write(*,*)m
    stop



    contains

    recursive subroutine msort(x)
    implicit none
    integer(8), intent(inout) :: x(:)
 
    integer(8) :: n, mid, i, j, k
    integer(8), allocatable :: tmp(:)
    
    n = size(x)
    
    if(n == 1) then
      return
    end if
 
    !
    ! 前半と後半に分けてソート
    ! 1~midとmid+1~nをそれぞれソート済みにする
    !
    mid = n / 2
    call msort(x(:mid))
    call msort(x(mid+1:))
 
    !
    ! tmpという配列にxを一時的に保管
    ! マージソートは外部ソート
    !
    allocate(tmp(n))
    tmp = x
 
    ! tmpの後半部分を左右反転
    do k = n, mid+1, -1
      tmp(k) = x(mid + n - k + 1)
    end do
 
    ! iは前半のリストのうちまだ見ていないものの最小値
    ! jは後半のリストのうちまだ見ていないものの最小値
    ! kは見た回数の合計
    i = 1
    j = n
 
    do k = 1, n
      !
      ! 一番左と一番右のうち小さい方をxに入れていく
      ! ここが等号つき不等号なので、安定ソートとなる
      !
      if(tmp(i) <= tmp(j)) then
        x(k) = tmp(i)
        i = i + 1
      else
        x(k) = tmp(j)
        j = j - 1
      end if
    end do
    
    deallocate(tmp)
    return
  end subroutine msort
 
    
    
end program sample