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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

type ary=array[1..100010] of int64;
var n,k,i:int64;
    d:ary;
procedure sort(p,q :int64; var d:ary);
  var i, j, k, w :int64;
begin i := p; j := q; k:=d[(p+q) div 2];
  repeat
    while d[i] < k do i := i+1;
    while k < d[j] do j := j-1;
    if i <= j then begin
      w := d[i]; d[i] := d[j]; d[j] := w;
      i := i+1;  j := j-1
    end
  until i > j;
  if p < j then sort(p,j,d);
  if i < q then sort(i,q,d)
end;
begin
read(n);
for i := 1 to n do read(d[i]);
sort(1,n,d);
write(d[(n div 2)+1]-d[n div 2])
end.