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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <bits/stdc++.h>
#include <iomanip>
#include <numeric>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (ll i = a; i < (ll)b; ++i)

#define rep(i, a, b) for (ll i = a; i < (ll)b; ++i)
#define INF 10e12
#define MAX 51000
#define all(x) (x).begin(), (x).end()
#define MX(x) *max_element(all(x))
#define MN(x) *min_element(all(x))

int main(void)
{
    ll n;
    cin >> n;
    priority_queue<ll, vector<ll>, greater<ll>> a;
    priority_queue<ll> b;
    vector<ll> vec(3 * n);
    rep(i, 0, 3 * n) cin >> vec[i];
    ll x = 0, y = 0, ans = -INF;
    rep(i, n, 2 * n)
    {
        x = 0, y = 0;
        rep(j, 0, n) a.push(vec[j]), x += vec[j];
        rep(j, n, i)
        {
            a.push(vec[j]);
            ll tmp = a.top();
            a.pop();
            x += vec[j] - tmp;
        }
        rep(j, 2 * n, 3 * n) b.push(vec[j]), y += vec[j];
        rep(j, i, 2 * n)
        {
            b.push(vec[j]);
            ll tmp = b.top();
            b.pop();
            y += vec[j] - tmp;
        }
        // cout << x << " " << y << endl;
        ans = max(ans, x - y);
    }
    cout << ans << endl;
}