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)

#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,fma,tune=native")
#include<bits/stdc++.h>
#define rall(x) x.rbegin(), x.rend()
#define all(x) x.begin(), x.end()
#define pb push_back
#include <x86intrin.h>
#define pb push_back
using namespace std;
using ll = long long;
using pi = pair<int, int>;
using vi = vector<ll>;
using vpi = vector<pi>;
using ld = long double;
const int maxn = 1e5 + 55, mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin >> n;
	vi a(n);
	for(auto &i : a) cin >> i;
	sort(all(a));
	cout << a[n/2] - a[n/2 - 1];
}