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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <list>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <chrono>
#include <math.h>
using namespace std;

using lli = long long int;
using Vint = std::vector<int>;
using Vlli = std::vector<lli>;
using Wint = std::vector<Vint>;
using Wlli = std::vector<Vlli>;
using Vbool = std::vector<bool>;
using Wbool = std::vector<Vbool>;
using pii = std::pair<int, int>;
using pll = std::pair<lli, lli>;
template <class T>
using Vec = std::vector<T>;
template <class T>
using Wec = Vec<Vec<T>>;

constexpr int MOD = 1e9 + 7;
constexpr int INFi = 2e9 + 1;
constexpr lli INFl = (lli)(9e18) + 1;
const std::vector<std::pair<int, int>> DXDY = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
constexpr char BR = '\n';

#define DEBUG(x) std::cerr << #x << " = " << x << '\n';
#define FOR(i, a, b) for(int (i) = (a); (i) < (b); ++(i))
#define FOReq(i, a, b) for(int (i) = (a); (i) <= (b); ++(i))
#define rFOR(i, a, b) for(int (i) = (b); (i) >= (a); --(i))
#define FORstep(i, a, b, step) for(int (i) = (a); i < (b); i += (step))
#define REP(i, n) FOR(i, 0, n)
#define rREP(i, n) rFOR(i, 0, (n-1))
#define vREP(ele, vec) for(auto &(ele) : (vec))
#define vREPcopy(ele, vec) for(auto (ele) : (vec))
#define SORT(A) std::sort((A).begin(), (A).end())
#define RSORT(A) std::sort((A).rbegin(), (A).rend())
#define ALL(A) (A).begin(), (A).end()
// 座標圧縮 (for vector) : ソートしてから使うのが一般的 ; SORT(A) => COORDINATE_COMPRESSION(A)
#define COORDINATE_COMPRESSION(A) (A).erase(unique((A).begin(),(A).end()),(A).end())



template <class T> inline int argmin(std::vector<T> vec){return min_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline int argmax(std::vector<T> vec){return max_element(vec.begin(), vec.end()) - vec.begin();}
template <class S, class T> inline void chmax(S &a, T b){if(a < b) a = b;}
template <class S, class T> inline void chmin(S &a, T b){if(a > b) a = b;}
template <class T> inline void reverseSORT(Vec<T> &Array){
  std::sort(Array.begin(), Array.end(), std::greater<T>());
}
inline int BitI(int k){return 1 << k;}
inline lli BitL(int k){return 1LL << k;}
inline void putsDouble(double d){printf("%.16lf\n", d);}
template <class T> inline std::string toString(T n){
  if(n == 0) return "0";
  std::string res;
  if(n < 0){n = -n;while(n != 0){res += (char)(n % 10 + '0'); n /= 10;}
  std::reverse(res.begin(), res.end()); return '-' + res;}
  while(n != 0){res += (char)(n % 10 + '0'); n /= 10;} std::reverse(res.begin(), res.end()); return res;
}

namespace MyFunc{
    using LLi = long long int;
    // GCD(a, b) ; a, bの最大公約数を求める関数
    inline LLi gcd(LLi a, LLi b){ while(b != 0){ a %= b; std::swap(a, b);} return a; }
    // LCM(a, b) ; a, bの最小公倍数を求める関数
    inline LLi lcm(LLi a, LLi b){ return (a * b) / MyFunc::gcd(a, b);}
    // 累乗を求める関数
    inline LLi power(LLi a, LLi n){
        LLi res = 1LL, waiting = a;
        while(n != 0LL){ if((n & 1LL) != 0LL) res *= waiting; waiting *= waiting; n >>= 1;}
        return res;
    }
    // 累乗の余りを求める関数
    inline LLi power_mod(LLi a, LLi n, LLi mod_number___ = 1e9 + 7){
        LLi res = 1LL, waiting = a;
        while(n != 0LL){ if((n & 1LL) != 0LL){ res *= waiting; res %= mod_number___;}
            waiting *= waiting; waiting %= mod_number___; n >>= 1; }
        return res;
    }
    // Z/pZ上の逆元を求める関数 (フェルマーの小定理)
    inline LLi inverse_mod(LLi a, LLi mod_number___ = 1e9 + 7){
        return MyFunc::power_mod(a, mod_number___-2);
    }
    inline LLi inverse_mod_euclid(LLi a, LLi mod_number___ = 1e9+7){
        LLi b = mod_number___, u = 1, v = 0;
        while (b != 0) { LLi t = a / b; a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);}
        u %= mod_number___; if (u < 0) u += mod_number___;
        return u;
    }
    // 素数であるかを判定する関数
    template <typename Integer_type>
    inline bool isPrime(Integer_type n){
        if(n < 2) return false; if(n == 2) return true; if(n % 2 == 0) return false;
        for(Integer_type x = 3; x * x <= n; ++++x) if(n % x == 0) return false;
        return true;
    }
    // 素数であるかの真偽表を返す　: n ≥ 1
    inline std::vector<bool> primeTable(int n){
        std::vector<bool> res(n+1, true); res[0] = false; res[1] = false;
        for(int x = 2; x * x <= n; ++x) if(res[x]){
            for(int i = 2 * x; i <= n; i += x){ res[i] = false; } }
        return std::move(res);
    }
    // 素因数分解したベクトルを返す　; {素因数, 指数}
    template <typename Integer_type>
    inline std::vector<std::pair<Integer_type, int>> prime_factorization(Integer_type n){
        std::vector<std::pair<Integer_type, int>> res(0);
        if(n <= 0) return std::move(res); // 例外処理　: nが 0 以下
        if(n % 2 == 0){
            n /= 2; int cnt = 1; while(n % 2 == 0){ n /= 2; cnt++;}
            res.emplace_back(make_pair(2, cnt));
        }
        Integer_type x = 3;
        while(x * x <= n){
            if(n % x == 0){
                n /= x; int cnt = 1; while(n % x == 0){ n /= x; cnt++; }
                res.emplace_back(make_pair(x, cnt));
            } ++++x;
        } if(n > 1) res.emplace_back(make_pair(n, 1));
        return std::move(res);
    }
}

// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~




int main(void){
    int n; scanf("%d", &n);
    Vint A(n); REP(i, n) scanf("%d", &A[i]); SORT(A);
    int k = n >> 1;
    printf("%d\n", A[k] - A[k-1]);
    return 0;
}