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 OCaml (4.02.3) (judge verdict: Wrong Answer)

open Batteries

let read_two () = Scanf.sscanf (read_line ()) "%d %d" (fun a b -> (a, b))

let read_three () =
  Scanf.sscanf (read_line ()) "%d %d %d" (fun a b c -> (a, b, c))

let read_four () =
  Scanf.sscanf (read_line ()) "%d %d %d" (fun a b c d -> (a, b, c, d))

let read_list () =
  read_line () |> String.split_on_char ' '
  |> List.map (fun s -> int_of_string s)

(* a**x *)
let pow a x =
  let rec aux acc = function 0 -> acc | n -> aux (acc * a) (n - 1) in
  aux 1 x

let rec pow2 n = if n mod 2 = 0 then 1 + pow2 (n / 2) else 0

let rec f n = function
  | d1 :: d2 :: rest -> if n = 1 then (d1, d2) else f (n - 2) rest
  | _ -> (0, 0)

let () =
  let n = read_int () in
  let d = read_list () |> List.sort compare in
  let a, b = f (n / 2) d in
  print_int (b - a);
  print_newline ()