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
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl "\n"
#define pb push_back
#define all(a) (a).begin(),(a).end()
#define rev(a) (a).rbegin(),(a).rend()
#define pll pair<ll,ll>
#define rep(i,a,b) for(long long i = (a); i <= (b); i++)

const int mod = 1e9+7;
const int N = 1e5+5; 

int main(){
	fast;
	ll n; cin >> n;
	ll arr[n+5];
	rep(i,1,n) cin >> arr[i];
	
	
	sort(arr + 1, arr + n + 1);
	
	
	ll f[N] = {};
	
	rep(i,1,n){
	    f[arr[i]]++;
	}
	
	rep(i,1,N-1){
	    f[i] += f[i-1];
	}
	
	ll ans = 0;
	
	rep(i,1,N-1){
	    if(f[i-1] == n - f[i-1]) ans++;   
	}
	
	cout << ans << endl;
}