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;;
#define ll long long
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;
int main(){
    int n; cin >> n;
    priority_queue<ll, vector<ll>, greater<ll>> f;
    vector<ll> m;
    priority_queue<ll> e;
    REP(i,n){
        ll t; cin >> t; f.push(t);
    }
    REP(i,n){
        ll t; cin >> t; m.push_back(t);
    }
    REP(i,n){
        ll t; cin >> t; e.push(t);
    }
    auto mf = m.begin();
    auto me = m.rbegin();
    REP(i,n){
        ll fdel = *mf - f.top();
        ll edel = e.top() - *me;
        if(fdel > 0 && edel > 0){
            if(fdel > edel){
                f.pop();
                f.push(*mf);
                mf++;
            }else{
                e.pop();
                e.push(*me);
                me++;
            }
        }else{
            if(fdel > edel){
                me++;
            }else{
                mf++;
            }
        }
    }
    ll ans = 0;
    REP(i,n){
        ans += f.top() - e.top();
        f.pop();
        e.pop();
    }
    cout << ans << endl;

}