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)

var i,n,j,res:longint;
    a:array[1..100000] of longint;
procedure sort(l,r:longint);
var i,j,x,y: longint;
begin
  i:=l; j:=r; x:=a[random(r-l+1)+l];
  repeat
    while a[i] < x do inc(i);
    while x < a[j] do dec(j);
    if not(i > j) then
      begin
        y:=a[i];
        a[i]:=a[j];
        a[j]:=y;
        inc(i);
        dec(j);
      end;
 until i > j;
 if l < j then sort(l,j);
 if i < r then sort(i,r);
end;
function find(l,r,x:longint):longint;
var  mid:longint;
begin
  while l <= r do
    begin
      mid:=(l+r) >> 1;
      if (a[mid] < x) then l:=mid + 1 else r:=mid -1;
    end;
  exit(l-1);
end;
begin
  readln(n);
  for i:=1 to n do read(a[i]);
  sort(1,n);
  for i:=1 to a[n] do if find(1,n,i) = n div 2 then inc(res);
  writeln(res)
end.