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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
#include <stdbool.h>
#define MOD 1000000007
#define END printf("\n");return 0;
#define QS09(how_data,data) qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_09);
#define QS90(how_data,data) qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_90);
#define I(a) long a;scanf("%ld",&a);
void input_array(long how_data,long *data);
void output_array(long how_data,long *data);
void input_array2(long first , long second , long data[][2]);
long get_random(long min, long max);
long factorial(long n);
long fibonacci(long n);
long qsort_09(const int *sys1 , const int *sys2);
long qsort_90(const int *sys1 , const int *sys2);
long sel_max(long a , long b);
long sel_min(long a , long b);
long array_max(long how_data,long *data);
long array_min(long how_data,long *data);
long can_DP(long how_data,long *data,long how_can,bool *can);
long array_sum(long how_data,long *data);
long Leven_dist(char *now , char *target);
void prime_fact(long target, long *data);
long get_digit(long target);
long ncr(long n , long r);
long npr(long n , long r);
long nhr(long n , long r);
 
long loop1,loop2,loop3,loop4,loop5,i_temp;
char c_temp;
 
int main(void){
    I(how_data);
    long data[how_data*3];
    input_array(how_data*3,data);
    QS90(how_data*3,data);
    long ans=0;
    for(loop1=how_data;loop1<how_data*2;loop1++){
        ans+=data[loop1];
        //printf("%ld\n",data[loop1]);
    }
    printf("%ld\n",ans);
    END;
    
    
    
    
}
 
 

void input_array(long how_data,long *data){
    long loop;
    for(loop=0;loop<how_data;loop++){
        scanf("%ld",&data[loop]);
    }
    return ;
}
void output_array(long how_data,long *data){
    long loop;
    for(loop=0;loop<how_data;loop++){
        printf("%ld\n",data[loop]);
    }
    return ;
}
 
void input_array2(long first,long second,long data[][2]){
    long loopA,loopB;
    for(loopA=0;loopA<first;loopA++){
        for(loopB=0;loopB<second;loopB++){
            scanf("%ld",&data[loopA][loopB]);
        }
    }
    return ;
}
 
long get_random(long min, long max){   //指定の範囲から乱数を1つ返すやつ
    //srand((unsigned int)time(0));   //現在時刻で疑似乱数初期化
    rand();rand();rand();rand();    //乱数を空打ち
    return (rand()%(max+1-min))+min;
}
 
 
long factorial(long n){//n!のMOD10^9+7を返すやつ
    unsigned long long int ret=1;
    for(long i=1;i<=n;i++)ret=(ret*i)%1000000007;
    return (long)ret;
}
 
long qsort_09(const int *sys1 , const int *sys2){ //小さいのが上にくるやつ
    //qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_09);
    if(*sys1<*sys2){
        return -1;
    }else if(*sys1==*sys2){
        return 0;
    }else{
        return 1;
    }
}
 
long qsort_90(const int *sys1 , const int *sys2){ //大きいのが上にくるやつ
    //qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_90);
    if(*sys1<*sys2){
        return 1;
    }else if(*sys1==*sys2){
        return 0;
    }else{
        return -1;
    }
}
 
long fibonacci(long n){
    switch(n){
        case 0:return 0;
        case 1:return 1;
        default :return fibonacci(n-1)+fibonacci(n-2);
    }
}
 
long sel_max(long a,long b){
    if(a>b)return a;
    return b;
}
 
long sel_min(long a,long b){
    if(a>b)return b;
    return a;
}
 
long can_DP(long how_data,long *data,long how_can,bool *can){//Typical DP Contest A
    //data内で組み合わせられる和をcanに0 or 1で入れる
    //返り値はパターン数
    long loopA,loopB;
    long ret=0;
    for(loopA=0;loopA<how_can;loopA++){//初期化
        can[loopA]=0;
    }
    can[0]=1;//絶対にありえる
    for(loopA=0;loopA<how_data;loopA++){
        for(loopB=how_can-1;loopB>=0;loopB--){
            if(can[loopB]==1 && loopB+data[loopA]<how_can){
                can[loopB+data[loopA]]=1;
            }
        }
    }
    for(loopA=0;loopA<how_can;loopA++){
        if(can[loopA]==1){
            ret++;
        }
    }
    return ret;
}
long array_max(long how_data,long *data){
    long loop;
    long ret=data[0];
    for(loop=0;loop<how_data;loop++){
        if(ret<data[loop])ret=data[loop];
    }
    return ret;
}
long array_min(long how_data,long *data){
    long loop;
    long ret=data[0];
    for(loop=0;loop<how_data;loop++){
        if(ret>data[loop])ret=data[loop];
    }
    return ret;
}
long array_sum(long how_data,long *data){
    long ret=0;
    long loop;
    for(loop=0;loop<how_data;loop++){
        ret+=data[loop];
    }
    return ret;
}
long Leven_dist(char *now , char *target){
    long loopA,loopB;
    //レーベンシュタイン距離を求める
    // 参考文献
    // http://nw.tsuda.ac.jp/class/algoB/c13.html  (アルゴリズム理解)
    // http://d.hatena.ne.jp/ohnishiakira/20090809/1249845529  (実装)
    long len_now=strlen(now)+1;
    long len_target=strlen(target)+1;
    long d[len_now][len_target]; //計算用
    for(loopA=0;loopA<len_now;loopA++) d[loopA][0]=loopA;
    for(loopA=0;loopA<len_target;loopA++) d[0][loopA]=loopA;
    for(loopA=1;loopA<len_now;loopA++){
        for(loopB=1;loopB<len_target;loopB++){
            long cost=(now[loopA-1]==target[loopB-1] ? 0:1);
            d[loopA][loopB]=sel_min(sel_min(d[loopA-1][loopB]+1,d[loopA][loopB-1]+1),d[loopA-1][loopB-1]+cost);
        }
    }
    return d[len_now-1][len_target-1];
}
void prime_fact(long target, long *data){
    long loopB=0;
    long loopA=2;
    long moto_target=target;
    while(target!=1){
        loopA-=1;
        while(1){
            loopA++;
            if(loopA>=sqrt(moto_target)+100){
                data[loopB]=target;
                target=1;
                break;
            }
            if(target%loopA==0){
                data[loopB]=loopA;
                target/=loopA;
                loopB++;
                break;
            }
        }
    }
    return ;
}
long get_digit(long target){
    return (long)(log10(target)+1);
}
long ncr(long n , long r){
    //パスカルの三角形
    long loopA,loopB;
    long pascal[100][102]={{0}};
    pascal[1][0]=1;
    pascal[1][1]=1;
    for(loopA=2;loopA<100;loopA++){
        pascal[loopA][0]=1;
        for(loopB=1;loopB<loopA;loopB++){
            pascal[loopA][loopB]=pascal[loopA-1][loopB-1]+pascal[loopA-1][loopB];
        }
        pascal[loopA][loopA]=1;
    }
    return pascal[n][r];
}
long npr(long n, long r){
    return ncr(n,r)*factorial(r);
}
long nhr(long n , long r){
    return ncr(n+r-1,r);
}