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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE BinaryLiterals #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE NumDecimals #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}

import Control.Applicative
import Control.Arrow
import Control.Monad
import Control.Monad.ST

import Data.Char
import Data.List
import Data.Maybe

import qualified Data.ByteString.Char8 as B

import Data.Array.ST.Safe
import Data.STRef

import Debug.Trace
import Text.Printf

readInt = readLn :: IO Int
readInts = map ( fst . fromJust . B.readInt ) . B.words <$> B.getLine
readNInts = readInt >>= flip replicateM readInt

which a b f = if f then a else b
mp [ a, b ] = ( a, b )

main = do
	n <- readInt
	ds <- sort <$> readInts
	let
		lb = last $ take ( n `div` 2 ) ds
		ub = head $ drop ( n `div` 2 ) ds
	print $ ub - lb