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;
#define ll                  long long int
#define ull                 unsigned long long int
#define ld                  long double
#define fi                  first
#define se                  second
#define pb                  push_back
#define pbb                 pop_back
#define mp                  make_pair
#define pii     	        pair<int,int>
#define all(x)              x.begin(), x.end()
#define uniq(x)             x.erase(unique(x.begin(),x.end()),x.end())
#define mem(array, value)   memset(array, value, sizeof(array))
#define PI      	        2*acos(0.0)
#define EPS     	        1e-8
#define line                cout << "\n==========\n"
#define fastRead 	        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define trace(...)          __f( #__VA_ARGS__ , __VA_ARGS__ )
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " : " << arg1 << " , ";
    __f(comma + 1, args...);
}

const int INF = 1e9+9;

const int nax = 1e5+5;
int ara[nax];

int main ()
{
    #ifdef Lollipop
        //freopen ("input.txt", "r", stdin);
        //freopen ("output.txt", "w", stdout);
    #endif

    int n;
    cin >> n;
    for (int i = 0; i < n; i++){
        cin >> ara[i];
    }
    sort(ara, ara+n);
    cout << ara[n/2] - ara[n/2-1] << endl;

    return 0;
}