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: Time Limit Exceeded)

program cut
  implicit none
  integer(8) :: n, ans
  integer(8),allocatable :: d(:)

  read *, n
  allocate(d(n))
  read *, d
  d = qsort(d)
  ans = d(n/2+1)-d(n/2)
  print '(i0)', ans
  stop

contains
  recursive function qsort(a) result(sorted)
    implicit none
    integer(8):: countl,countg, pivot, i, n
    integer(8),intent(in):: a(:)
    integer(8),allocatable:: sorted(:),less(:),great(:)

    n = size(a)
    allocate(sorted(n))
    if (n <= 1) then
      sorted = a
    else
      sorted = 0
      countl = 1
      countg = n
      pivot = a(1)
      do i =2, n
        if (pivot >= a(i)) then
          sorted(countl) = a(i)
          countl = countl + 1
        else
          sorted(countg) = a(i)
          countg = countg - 1
        end if
      end do
      allocate(less(countl-1),great(n-countg))
      less = sorted(1:countl - 1)
      great = sorted(countg+1:n)
      sorted(1:countl-1) = qsort(less)
      sorted(countl) = pivot
      sorted(countg+1:n) = qsort(great)
    end if
  end function
end program cut