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 5.4.1) (judge verdict: Accepted)

#include <float.h>
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_ARRAY_MAX		300000									// 最大配列数

// 内部変数
static FILE *szpFpI;											// 入力
static int si1Array[D_ARRAY_MAX];								// 配列
static int siACnt;												// 配列数

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpA;
	static int siTNo;
#endif

// １行出力
int
fOutLine(
	char *pcpLine				// <I> １行
)
{
	char lc1Buf[1024];

#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, pcpLine)) {
		siRes = -1;
	}
#else
	printf("%s", pcpLine);
#endif

	return 0;
}

// ソート関数 - int降順
int
fSortFncID(
	const void *pzpVal1			// <I> 値１
	, const void *pzpVal2		// <I> 値２
)
{
	int *lipVal1 = (int *)pzpVal1;
	int *lipVal2 = (int *)pzpVal2;

	// int降順
	if (*lipVal1 > *lipVal2) {
		return -1;
	}
	else if (*lipVal1 < *lipVal2) {
		return 1;
	}

	return 0;
}

// 実行メイン
long long
fMain(
)
{
	int i;
	char lc1Buf[1024];

	// 配列数 - 取得
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	sscanf(lc1Buf, "%d", &siACnt);

	// 配列 - 取得
	for (i = 0; i < siACnt * 3; i++) {
		fscanf(szpFpI, "%d", &si1Array[i]);
	}
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	qsort(si1Array, siACnt * 3, sizeof(int), fSortFncID);

	// 和 - 取得
	long long llSum = 0;
	for (i = 0; i < siACnt; i++) {
		llSum += si1Array[i * 2 + 1];
	}

	return llSum;
}

// １回実行
int
fOne(
)
{
	long long llRet;
	char lc1Buf[1024];

	// 入力 - セット
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", siTNo);
	szpFpI = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", siTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#else
	szpFpI = stdin;
#endif

	// 実行メイン
	llRet = fMain();

	// １行出力
	sprintf(lc1Buf, "%lld\n", llRet);
	fOutLine(lc1Buf);

	// 残データ有無
#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, "")) {
		siRes = -1;
	}
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpI);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", siTNo);
	}
	else {
		printf("NG %d\n", siTNo);
	}
#endif

	return 0;
}

// プログラム開始
int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		siTNo = i;
		fOne();
	}
#else
	fOne();
#endif

	return 0;
}