Problem:
Score : 500 points

Problem Statement

Let N be a positive integer.

There is a numerical sequence of length 3N, a = (a_1, a_2, ..., a_{3N}).
Snuke is constructing a new sequence of length 2N, a', by removing exactly N elements from a without changing the order of the remaining elements.
Here, the score of a' is defined as follows: (the sum of the elements in the first half of a') - (the sum of the elements in the second half of a').

Find the maximum possible score of a'.

Constraints

1 ≤ N ≤ 10^5

a_i is an integer.

1 ≤ a_i ≤ 10^9

Partial Score

In the test set worth 300 points, N ≤ 1000.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{3N}

Output

Print the maximum possible score of a'.

Sample Input 1

2
3 1 4 1 5 9

Sample Output 1

1

When a_2 and a_6 are removed, a' will be (3, 4, 1, 5), which has a score of (3 + 4) - (1 + 5) = 1.

Sample Input 2

1
1 2 3

Sample Output 2

-1

For example, when a_1 are removed, a' will be (2, 3), which has a score of 2 - 3 = -1.

Sample Input 3

3
8 2 2 7 4 6 5 3 8

Sample Output 3

5

For example, when a_2, a_3 and a_9 are removed, a' will be (8, 7, 4, 6, 5, 3), which has a score of (8 + 7 + 4) - (6 + 5 + 3) = 5.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

int main(){
    int n, i, a, b, c, s=0, t=0;
    std::vector<int> v;
    cin>>n;
    for(i=1; i<=(n*3); i++){
        cin>>a;
        v.push_back(a);
    }
    if(n%2==0){
        b=n*3;
        b/=2;
    }
    else{
        b=n*3;
        b/=2;
        b+=1;
    }
    c=v[b];
    for(i=1; i<b; i++){
        if(v[i-1]>=c){
            t++;
        }
    }
    if(t<n){
        sort(v.begin(), v.begin()+b);
        for(i=1; i<=n; i++){
            s+=v[b-i];
        }
        sort(v.begin()+b+1, v.end());
        for(i=0; i<n; i++){
            s-=v[b+i];
        }
    }
    else{
        sort(v.begin(), v.begin()+b-1);
        for(i=1; i<=n; i++){
            s+=v[b-i];
        }
        sort(v.begin()+b, v.end());
        for(i=0; i<n; i++){
            s-=v[b+i];
        }
    }
    cout<<s;
    return 0;
}