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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

// 総数を1000000007（素数）で割った余り
const long long mod = 1e9 + 7;

using ll = long long;
using pii  = pair<int, int>;
using pll = pair<ll, ll>;
#define ull unsigned long long
#define ld long double
#define vi vector<int>
#define vll vector<ll>
#define vc vector<char>
#define vs vector<string>
#define vpii vector<pii>
#define vpll vector<pll>

#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; i++)
#define rep1(i, n) for (int i = 1, i##_len = (n); i <= i##_len; i++)
#define repr(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rep1r(i, n) for (int i = ((int)(n)); i >= 1; i--)

#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define RSORT(x) sort(rall(x));
#define pb push_back
#define mp make_pair

#define INF (1e9)
#define PI (acos(-1))
#define EPS (1e-7)

ull gcd(ull a, ull b) { return b ? gcd(b, a % b) : a; }
ull lcm(ull a, ull b) { return a / gcd(a, b) * b; }

int main(){
    int n;
    cin >> n;
    vi d(n);
    rep(i, n) cin >> d[i];
    sort(d.begin(), d.end());
    int ans = d[n/2] - d[n/2-1];
    cout << ans << endl;
    return 0;
}