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: Wrong Answer)

module UIntegerSort
  implicit none
 
  private
 
  public QuickSortInteger
 
contains
    
  !!
  ! ピボット値の取得
  !
  subroutine GetPivotIntegerValue(array, n, a, b, pivot_value)
    integer, intent(inout) :: array(:), pivot_value
    integer, intent(in) :: n, a, b
 
    pivot_value = array((a + b) / 2)
  end subroutine
 
  !!
  ! ピボット値に従って値を分離
  !
  subroutine GetPartitionInteger(array, n, a, b, pivot)
    integer, intent(inout) :: array(:)
    integer, intent(in)  :: n, a, b
    integer, intent(out) :: pivot
 
    integer :: buffer, pivot_value
    integer :: l, r
 
    if (a == b) then
      pivot = 0
      return
    end if
 
    call GetPivotIntegerValue(array, n, a, b, pivot_value)
 
    l = a
    r = b
 
    do while(.true.)
      do while(array(l) < pivot_value)
        if (l >= b) exit
        l = l + 1
      end do
 
      do while(pivot_value < array(r))
        if (r <= a) exit
        r = r - 1
      end do
 
      if(r <= l) exit
 
      buffer   = array(l)
      array(l) = array(r)
      array(r) = buffer
 
      l = l + 1
      r = r - 1
    end do
 
    pivot = r
  end subroutine
 
  !!
  ! 配列の一部分をソートする。
  !
  recursive subroutine QuickSortIntegerSubarray(array, n, a, b)
    integer, intent(inout) :: array(:)
    integer, intent(in)    :: n, a, b
 
    integer :: pivot
 
    call GetPartitionInteger(array, n, a, b, pivot)
    if (pivot /= 0) then
      call QuickSortIntegerSubarray(array, n, a, pivot)
      call QuickSortIntegerSubarray(array, n, pivot + 1, b)
    end if
  end subroutine
 
  !!
  ! 配列全体をソートする
  !
  subroutine QuickSortInteger(array, n)
    integer :: array(:)
    integer :: n
 
    call QuickSortIntegerSubarray(array, n, 1, n)
  end subroutine
 
end module
 
program TestIntegerSort
  use UIntegerSort
 
  implicit none
  integer,allocatable::a(:)
  integer :: N,i
  read*,N
  allocate(a(N))
  read(*,*)(a(i),i=1,N)
  call QuickSortInteger(a, N)
    print *, a(N/2+1)-a(N)
end program