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)

Program Catcoder1;
//uses crt;
var A:array[1..1000000] of longint;
    i,n,s:longint;
//--------------------------------
procedure QS(l,r :longint);
var i,j,x,temp :longint;
begin
i:=l;
j:=r;
x:=A[(l+r) div 2];
repeat
While A[i]<x do inc(i);
While A[j]>x do dec(j);
if i<=j then begin
             temp:=A[i];
             A[i]:=A[j];
             A[j]:=temp;
             inc(i);
             dec(j);
             end;
until i>j;
if i<r then QS(i,r);
if j>l then QS(l,j);
end;
//------------------------------------
begin
readln(n);
for i:=1 to n do begin
                 read(A[i]);
                 end;
QS(1,n);
s:=A[(n div 2)+1]-A[n div 2];
Writeln(s);
end.