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: Accepted)

#include <bits/stdc++.h>

using namespace std;


#ifdef zxc

#include <sys/resource.h>
#include "debug.h"
#include "profile.h"

#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define profile(X) LOG_DURATION(X)
#else
#define debug(...) 42
#define profile(...) 42
#endif

using ll = int64_t;
using ld = double;
const ld EPS = 1e-9;
//const ll MOD = 924844033;
const ld PI = 3.141592653589793;
const int maxn = 600001;
const ll MOD = 1e9 + 7;
using cd = complex<double>;

void fast_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
}

signed main() {
#ifdef zxc
    struct rlimit rl;
    const rlim_t kStackSize = 512L * 1024L * 1024L;
    assert(!getrlimit(RLIMIT_STACK, &rl));
    rl.rlim_cur = kStackSize;
    assert(!setrlimit(RLIMIT_STACK, &rl));
    freopen("../input.txt", "r", stdin);
//    freopen("../kolya_output.txt", "w", stdout);
#else
    //    mt19937 mt(std::chrono::high_resolution_clock::now().time_since_epoch().count());
    //            freopen("", "r", stdin);
    //        freopen("hack.out", "w", stdout);
#endif
    profile("all");
    auto solve = [](int _) {
        set<ll> s;
        int n;
        cin >> n;
        vector<ll> a(3 * n + 1);
        for (int i = 1; i <= 3 * n; ++i) {
            cin >> a[i];
        }
        priority_queue<ll> pq;
        vector<ll> dp_suf(3 * n + 1);
        vector<ll> dp_pref(3 * n + 1);
        ll sum = 0;
        for (int i = 3 * n; i >= 1; --i) {
            pq.push(a[i]);
            sum += a[i];
            if (pq.size() > n) {
                sum -= pq.top();
                pq.pop();
            }
            if (pq.size() == n) {
                dp_suf[i] = sum;
            }
        }
        priority_queue<ll, vector<ll>, greater<>> pq2;
        sum = 0;
        for (int i = 1; i <= 3 * n; ++i) {
            sum += a[i];
            pq2.push(a[i]);
            if (pq2.size() > n) {
                sum -= pq2.top();
                pq2.pop();
            }
            if (pq2.size() == n) {
                dp_pref[i] = sum;
            }
        }
        ll ans = -1e18;
        for (int i = n; i <= 2 * n; ++i) {
            ans = max(ans, dp_pref[i] - dp_suf[i + 1]);
        }
        cout << ans;
    };


    fast_io();
    cout.precision(9);
    cout << fixed;

    solve(1);


}