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
	N,i:Longint;
	d:Array[1..100000]of Longint;
function bs(x:Longint):Longint;
var L,R,M,cnt,i:Longint;
begin
	L:=0;
	R:=100000;
	while R-L>1 do begin
		M:=(L+R)div 2;
		cnt:=0;
		for i:=1 to N do if d[i]<=M then inc(cnt);
		if cnt>=x then R:=M else L:=M;
	end;
	bs:=R;
end;
begin
	read(N);
	for i:=1 to N do read(d[i]);
	writeln(bs(N div 2+1)-bs(N div 2));
end.