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

typedef long long ll;
const ll INF = 1e9, MOD = 1e9 + 7;
const double EPS = 1e-9, PI = 3.141592653589793;
#define Cin(x) cin >> x
#define Cin2(x,y) cin >> x >> y
#define Cin3(x,y,z) cin >> x >> y >> z
#define Cin4(w,x,y,z) cin >> w >> x >> y >> z
#define Cout(x) cout << x << "\n"
#define Cout2(x,y) cout << x << y << "\n"
#define Cout3(x,y,z) cout << x << y << z << "\n"
#define Cout4(w,x,y,z) cout << w << x << y << z << "\n"
#define Kaigyou cout << "\n"
#define REP(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)
#define REPR(i,a,b) for(ll i=b;i>=a;i--)
#define repr(i,n) REPR(i,0,n)
#define Size(n) (n).size()
#define All(v) (v).begin(), (v).end()

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll n,a[300001],ans=0; Cin(n);
    rep(i,3*n) Cin(a[i]);
    sort(a,a+3*n,greater<int>());
    for(int i=1;i<2*n;i+=2){
        ans += a[i];
    }
    Cout(ans);
    return 0;
}