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 (GCC 9.2.1) (judge verdict: Accepted)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define _USE_MATH_DEFINES
#include <math.h>
#include <float.h>

#pragma region 定義
typedef long long  ll;
typedef unsigned long long  ull;


#define _itoa(A,N) sprintf(A, "%d", N);

int i_dsort(const void* a, const void* b);
int i_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);

int s_asort(const void* a, const void* b);
int s_dsort(const void* a, const void* b);

//昇順
int s_asort(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(strcmp((char*)b, (char*)a));
}

int i_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int i_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}


int i_gcd(int a, int b);
int i_gcd(int a, int b) {
	if (!b)return a;
	return i_gcd(b, a % b);
}
int i_lcm(int a, int b);
int i_lcm(int a, int b) {
	return (a * b) / i_gcd(a, b);
}
int ll_gcd(ll a, ll b);
int ll_gcd(ll a, ll b) {
	if (!b)return a;
	return ll_gcd(b, a % b);
}
int ll_lcm(ll a, ll b);
int ll_lcm(ll a, ll b) {
	return (a * b) / ll_gcd(a, b);
}

// puts("Yes\n") puts("No\n")
// scanf("%d.%d") 浮動小数点入力を整数２つで得る


//#define pai  3.1415926535 
//#define  _USE_MATH_DEFINES でM_PIが使える

#define deg_rad(X)  (X*(M_PI/180))


int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}
#define _abs(x) ((x)>0?(x):-(x))
#define _min(x,y) ((x)<(y)?(x):(y))
#define _max(x,y) ((x)<(y)?(y):(x))

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

//素因数分解　pf素数が入る　pc 素数の数
//ll _pfact(ll a);
//static ll pf[100000000];
//static ll pc;
//
//ll _pfact(ll a) {
//	long i, r;
//	if (a % 2 == 0) {
//		pf[pc] = 2; pc++;
//		return(_pfact(a / 2));
//	}
//	r = sqrt(a);
//	for (i = 3; i <= r; i += 2) {
//		if (a % i == 0) {
//			pf[pc] = i; pc++;
//			return(_pfact(a / i));
//		}
//	}
//	if (a != 1) {
//		pf[pc] = a; pc++;
//	}
//	return 0;
//
//}

//同じものの最初の文字をNULLにして降順
int _sort(const void* a, const void* b);
int _sort(const void* a, const void* b) {
	char rc;

	rc = strcmp((char*)b, (char*)a);
	if (rc == 0) {
		*(char*)a = 0x00;
		return(1);
	}
	return(rc);
}
// 立っている ビットの数
int bitcount(unsigned int uc);
int bitcount(unsigned int uc)
{
	int bit;

	for (bit = 0; 0 != uc; uc = (unsigned int)(uc >> 1))
		if (0 != (uc & 1))++bit;

	return bit;
}//bitco

//10進数の桁数の総和
int _sumdig(int n);
int _sumdig(int n) {
	int sum = 0;
	while (n > 0) {	sum += n % 10;n /= 10;}
	return(sum);
}



#define _rnd(a,b)    ((a+b-1)/b) //　A/Bの余りを繰り上げる
#define _abs(x) ((x)<0 ? -(x) : (x))
#define _MAX 200000
#pragma endregion

// ABC145 C - Average Length
//int main(void) {
//	int n,i, x[9],y[9],cnt=0;
//	double  sum=0;
//
//	scanf("%d", &n);
//	for (i = 0; i < n; i++) 
//		scanf("%d %d", &x[i], &y[i]);
//
//		for(i=0;i<n-1;i++)
//			for (int j = i + 1; j < n; j++) {
//				sum += hypot(x[i] - x[j], y[i] - y[j]);
//				cnt++;
//				printf("%d %lf\n", cnt, sum);
//			}
//	printf("%lf\n",sum/cnt);
//	return 0;
//}

// AGC012 A - AtCoder Group Contest
static int a[300001];
int main(void) {
	int n, i;

	scanf("%d", &n); n *= 3;
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);

	qsort(a, n, sizeof(int), i_dsort);

	ll sum = 0; ; n = n / 3 * 2;
	for (i = 1; i < n; i+=2)
		sum += a[i];

	printf("%lld", sum);
	return 0;
}