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 Go (1.6) (judge verdict: Runtime Error)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
	"strings"
)

var sc = bufio.NewScanner(os.Stdin)

// 一行読み出し
func nextLine() string {
	sc.Scan()
	return sc.Text()
}

// 文字列リバース関数
func strReverse(s string) string {
	rs := []rune(s)
	for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
		rs[i], rs[j] = rs[j], rs[i]
	}
	return string(rs)
}

// スペース区切りの文字列から int のスライスを得る
func intSlice(s string) []int {
	ss := strings.Split(s, " ")
	res := make([]int, len(ss))
	for i, v := range ss {
		iv, _ := strconv.Atoi(v)
		res[i] = iv
	}

	return res
}

func main() {

	var n int
	fmt.Scan(&n)

	s := nextLine()
	ns := intSlice(s)

	sort.Ints(ns)

	ans := 0
	if ns[n/2-1] == ns[n/2] {
		// 真ん中の2つの値が同じ場合等しくわけられない
		ans = 0
	} else {
		ans = ns[n/2] - ns[n/2-1]
	}
	fmt.Println(ans)
}