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 main
	implicit none
    integer::N,i,ans,r,b,k
    integer, allocatable :: a(:)
    read(*,*) N
    allocate(a(N))
    read(*, *)(a(i), i = 1,N)
    ans=0
	r=0
    b=0
    k=1
    do while(r>=b)
		r=0
        b=0
    	do i=1,N
        	if(a(i)<k) then
            	b=b+1
                r=r
            else 
                b=b
                r=r+1
            end if
       end do
       if(b==r) then 
       	ans=ans+1
       else 
        ans=ans
       end if
       k=k+1
   end do
  print*,ans
 end program main