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 Ruby (2.3.3) (judge verdict: Accepted)

# util functions

# debug
$debugFlag = ( ARGV[0] != nil )
p "==== DEBUG MODE ====" if $debugFlag

def pd(str)
  p str if $debugFlag
end

def printd(str)
  print str if $debugFlag
end

# input
inputs = []
lc = 0
while l = $stdin.gets do
  if false
    inputs << l.chomp.split(" ").map(&:to_s)
  else
    inputs << l.chomp.split(" ").map(&:to_i)
  end
  lc += 1
end

nn = inputs[0][0]
lst = inputs[1].sort

# input check
if $debugFlag
  p "--- INPUT (begin) ---"
  inputs.each do |l_i|
    p l_i
  end
  p "--- INPUT (end) ---"
  print "\n"
end

# body
if $debugFlag
  p nn
  p lst
end

idx = nn/2 - 1
pd idx
pd lst[idx+1]
pd lst[idx]

puts lst[idx+1] - lst[idx]