Problem:
Score : 300 points

Problem Statement

There are 3N participants in AtCoder Group Contest.
The strength of the i-th participant is represented by an integer a_i.
They will form N teams, each consisting of three participants.
No participant may belong to multiple teams.

The strength of a team is defined as the second largest strength among its members.
For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.

Find the maximum possible sum of the strengths of N teams.

Constraints

1 ≤ N ≤ 10^5

1 ≤ a_i ≤ 10^{9}

a_i are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{3N}

Output

Print the answer.

Sample Input 1

2
5 2 8 5 1 5

Sample Output 1

10

The following is one formation of teams that maximizes the sum of the strengths of teams:

Team 1: consists of the first, fourth and fifth participants.

Team 2: consists of the second, third and sixth participants.

Sample Input 2

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 2

10000000000

The sum of the strengths can be quite large.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

//def
#define debug(x) cout << #x << ": " << x << '\n'
#define out(x) cout << x << '\n'
#define repeat(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define revrepeat(i, a, b) for (int i = (int)(b)-1; i >= (int)(a); i--)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define revrep(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define foreach(e, v) for (auto &e : v)
#define all(x) (x).begin(), (x).end()
#define CYES cout << "Yes" << endl
#define CNO cout << "No" << endl
#define SPC(x) cout << fixed << setprecision(x)
#define ZERO(a) memset(a, 0, sizeof(a))
#define MINUS(a) memset(a, 0xff, sizeof(a))

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vector<int>> vii;
typedef vector<ll> vl;
typedef vector<vector<ll>> vll;
typedef pair<int, int> P;
constexpr int MOD = 1e9 + 7;
constexpr int INF = __INT_MAX__;               // 2^31 - 1
constexpr long long INFLL = __LONG_LONG_MAX__; // 2^61 - 1
constexpr int MAX_N = 1e5 + 5;
constexpr double PI = acos(-1);

template <class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

long long fac[MAX_N], finv[MAX_N], inv[MAX_N];
// テーブルを作る前処理 from: https://qiita.com/drken/items/3b4fdf0a78e7a138cd9a
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX_N; i++) {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k) {
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
// factorial
long long fact(int n) {
    if (n == 1) return fac[n] = 1;
    if (fac[n]) return fac[n];
    return fac[n] = (fact(n - 1) * n) % MOD;
}

long long modpow(long long m, long long p) {
    if (p == 0) return 1;
    if (p % 2)
        return m * modpow(m, p - 1) % MOD;
    else {
        long res = modpow(m, p / 2);
        return res * res % MOD;
    }
}
// https://qiita.com/drken/items/cce6fc5c579051e64fab
// UnionFind<int> uf(n);
template <class Abel>
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    vector<int> size;
    vector<Abel> diff_weight;

    UnionFind(int n = 1, Abel SUM_UNITY = 0) {
        init(n, SUM_UNITY);
    }

    void init(int n = 1, Abel SUM_UNITY = 0) {
        par.resize(n);
        rank.resize(n);
        size.resize(n);
        diff_weight.resize(n);
        for (int i = 0; i < n; ++i)
            par[i] = i, rank[i] = 0, size[i] = 1, diff_weight[i] = SUM_UNITY;
    }

    int root(int x) {
        if (par[x] == x) {
            return x;
        } else {
            int r = root(par[x]);
            diff_weight[x] += diff_weight[par[x]]; // path compression
            return par[x] = r;
        }
    }
    // returns size of connected-tree number
    int get_size(int x) {
        return size[root(x)];
    }

    Abel weight(int x) {
        root(x);
        return diff_weight[x];
    }

    bool is_same(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y, Abel w) {
        w += weight(x);
        w -= weight(y);
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y), w = -w;
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        size[x] = size[y] = size[x] + size[y];
        diff_weight[y] = w;
        return true;
    }

    Abel diff(int x, int y) {
        return weight(y) - weight(x);
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vl a(3 * n);
    rep(i, 3 * n) cin >> a[i];

    sort(all(a), greater<ll>());
    ll ans_a = 0;
    ll ans_b = 0;
    rep(i, n) ans_a += a.at(2 * (i + 1));
    repeat(i, n, 2 * n) ans_b += a[i];

    ll ans = max(ans_a, ans_b);
    out(ans);
}