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: Accepted)


#include <sstream>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <queue>
#include <bitset>
using namespace std;

#define int long long
#define FOR(i, a, b) for (int i = (a), _##i = (b); i <= _##i; ++i)
#define FORD(i, a, b) for (int i = (a), _##i = (b); i >= _##i; --i)
#define REP(i, a) for (int i = 0, _##i = (a); i < _##i; ++i)
#define REPD(i,n) for(int i = (n)-1; i >= 0; --i)

#define DEBUG(X) { cerr << #X << " = " << (X) << endl; }
#define PR(A, n) { cerr << #A << " = "; FOR(_, 1, n) cerr << A[_] << ' '; cerr << endl; }
#define PR0(A, n) { cerr << #A << " = "; REP(_, n) cerr << A[_] << ' '; cerr << endl; }

#define sqr(x) ((x) * (x))
#define ll long long
#define double long double
typedef pair<int, int> II;
#define PI (2 * acos((double)0))
#define __builtin_popcount __builtin_popcountll
#define SZ(x) ((int)(x).size())
#define ALL(a) (a).begin(), (a).end()
#define MS(a,x) memset(a, x, sizeof(a))
#define next ackjalscjaowjico
#define prev ajcsoua0wucckjsl
#define y1 alkscj9u20cjeijc
#define left lajcljascjljl
#define right aucouasocjolkjl
#define y0 u9cqu3jioajc

#define TWO(X) (1LL<<(X))
#define CONTAIN(S,X) ((S) & TWO(X))

double safe_sqrt(double x) { return sqrt(max((double)0.0, x)); }
int GI(int& x) { return scanf("%lld", &x); }

int n, a[300111];

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << (fixed) << setprecision(9) << boolalpha;
    while (cin >> n) {
        n *= 3;
        FOR(i,1,n) cin >> a[i];
        sort(a+1, a+n+1);
        int res = 0;
        int i = n-1;
        REP(turn,n/3) {
            res += a[i];
            i -= 2;
        }
        cout << res << endl;
    }
    return 0;
}