Niuniu's team competition

【 subject 】

【 Team competition 】

Niuniu held a programming competition , The participants in the competition are 3*n A contestant , Each player has a level value a_i. Now we're going to group these players , It consists of n A team , That is, each team 3 people . Niuniu found The level value of the team is equal to the second highest level value among the team members .

for example : The level values of three members of a team are 3,3,3. So the level of the team is 3 The level values of three members of a team are 3,2,3. So the level of the team is 3 The level values of three members of a team are 1,5,2. So the level of the team is 2 In order to make the game more interesting , Niuniu wants to arrange teams to maximize the sum of level values of all teams . As the example shows : If Niuniu put 6 Two players are divided into two teams , If the scheme is : team1:{1,2,5}, team2:{5,5,8}, At this time, the sum of the horizontal values is 7. And if the plan is : team1:{2,5,8}, team2:{1,5,5}, At this time, the sum of the horizontal values is 10. There is nothing like the sum of 10 A bigger plan , So the output 10.

Input description ：

The first line of input is a positive integer n(1 ≤ n ≤ 10^5) The second line includes 3*n It's an integer a_i(1 ≤ a_i ≤ 10^9), Represents the level value of each contestant .

Output description ：

Output an integer representing the sum of the level values of all teams, the maximum value .

Example 1：

Input ：

2

5 2 8 5 1 5

Output ：

10

【 Ideas 】

Rank all horizontal values in descending order first , Then take two numbers from the first position , These two numbers are respectively the highest level and the middle level in a group , Total n Group .

【 Code 】

#include<iostream>
#include<algorithm>
using namespace std;

bool cmp(int a,int b)
{
    
	return a>b;
}

int main()
{
    
	long long n;
	long long ans=0;
	cin>>n;
	long long* arr=new long long[3*n];
	for(long long i=0;i<3*n;i++) 
      cin>>arr[i];
	sort(arr,arr+3*n,cmp);
	for(long long i=1;i<2*n;i+=2)
      ans+=arr[i];
	cout<<ans<<endl;
	return 0;
}