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> 
#define int long long
#define quickie ios_base::sync_with_stdio(false); cin.tie(NULL);
#define rep(i, a, b) for(int i=a; i<b; i++)
#define rep1(i, a, b) for(int i=a; i<=b; i++)
#define repp(i, a, b) for(int i=b-1; i>=a; i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define db double
#define mi map<int, int>
#define mc map<char, int>
#define ms map<string, int>
#define vi vector<int>
#define qi queue<int>
#define MI(x) power(x, mod-2)
#define test int t; cin >> t;
//#define mod 1000000007LL
#define mod 998244353 
using namespace std;
int power(int x, int y) ;
int gcd(int a, int b) ;

signed main() {
    quickie
    int n ;
    cin >> n ;
    int a[n] ;
    rep(i ,0, n) cin >> a[i] ;
    sort(a, a+n) ;
    if(a[n/2] == a[n/2-1]) cout << 0 << "\n" ;
    else cout << (a[n/2] - a[n/2-1]) << "\n" ;
}

int power(int x, int y) {
    int res = 1; x %= mod;
    while (y > 0) {
        if (y & 1)
            res = (res*x) % mod;
        y = y>>1;
        x = (x*x) % mod;
    }
    return res%mod;
}
 
int gcd(int a,int b){
    if(a==0) return b;
    return gcd(b%a,a);
}