Problem:
Score : 300 points

Problem Statement

Takahashi made N problems for competitive programming.
The problems are numbered 1 to N, and the difficulty of Problem i is represented as an integer d_i (the higher, the harder).

He is dividing the problems into two categories by choosing an integer K, as follows:

A problem with difficulty K or higher will be for ARCs.

A problem with difficulty lower than K will be for ABCs.

How many choices of the integer K make the number of problems for ARCs and the number of problems for ABCs the same?

Problem Statement

2 \leq N \leq 10^5

N is an even number.

1 \leq d_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the number of choices of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 1

6
9 1 4 4 6 7

Sample Output 1

2

If we choose K=5 or 6, Problem 1, 5, and 6 will be for ARCs, Problem 2, 3, and 4 will be for ABCs, and the objective is achieved.
Thus, the answer is 2.

Sample Input 2

8
9 1 14 5 5 4 4 14

Sample Output 2

0

There may be no choice of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 3

14
99592 10342 29105 78532 83018 11639 92015 77204 30914 21912 34519 80835 100000 1

Sample Output 3

42685

Solution in Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program abc132c
  implicit none

  integer :: n, diff
  integer, allocatable :: d(:)

  read(*,*) n
  allocate( d(n) )
  read(*,*) d(:)

  call heapsort(n, d(:))

  diff =  d(n/2+1) - d(n/2)
  print *, diff
  deallocate( d )
  stop

contains

  subroutine heapsort(n,array)
    implicit none
    integer,intent(in) :: n
    integer,intent(inout) :: array(1:n)

    integer ::i,k,j,l
    integer  :: t

    if(n.le.0)then
       write(6,*)"Error, at heapsort"; stop
    endif
    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

end program abc132c