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 ABC170D
    implicit none
    integer(8)::N,i,j,ans=0,amax,as,flg
    integer(8),allocatable::A(:),B(:)
    read*,N
    allocate(A(N))
    read*,A
    amax=maxval(A)
    allocate(B(amax))
    B=0
    call heapsort(N,A)
    do i=1,N
        B(A(i))=1
    end do

    flg=0
    do i=1,N-1
        if(A(i)/=A(i+1).and.flg==1)then
            flg=0
            A(i)=0
        else if(A(i)==A(i+1))then
            flg=1
            A(i)=0
            if(i==N-1)A(N)=0
        end if
    end do

    do i=1,N
        if(A(i)/=0)then
            as=int(sqrt(real(A(i))))
            flg=0
            do j=2,as
                if(mod(A(i),j)==0)then
                    if(B(j)==1.or.B(A(i)/j)==1)then
                        flg=1
                        exit
                    end if
                end if
            end do
            if(flg==0)ans=ans+1
        end if
    end do

    print'(i0)',ans
end program ABC170D

subroutine heapsort(n,array)
    implicit none
    integer(8),intent(in) :: n
    integer(8),intent(inout) :: array(1:n)
   
    integer(8) ::i,k,j,l
    integer(8) :: t
   
    if(n.eq.1)return
  
    l=n/2+1
    k=n
    do while(k.ne.1)
       if(l.gt.1)then
          l=l-1
          t=array(L)
       else
          t=array(k)
          array(k)=array(1)
          k=k-1
          if(k.eq.1) then
             array(1)=t
             exit
          endif
       endif
       i=l
       j=l+l
       do while(j.le.k)
          if(j.lt.k)then
             if(array(j).lt.array(j+1))j=j+1
          endif
          if (t.lt.array(j))then
             array(i)=array(j)
             i=j
             j=j+j
          else
             j=k+1
          endif
       enddo
       array(i)=t
    enddo
  
    return
end subroutine heapsort