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 <iostream>
#include <queue>
#include <vector>
#include <functional>
using namespace std;

using ll = long long;

int main() {
    int N;
    cin >> N;

    const int MAXN = 200000;
    static ll a[MAXN * 3];

    for (int i = 0; i < 3 * N; i++) {
        cin >> a[i];
    }

    static ll sum1[MAXN * 3],   // 前半部分 [0, i)
              sum2[MAXN * 3];   // 後半部分 [i, 3N)
    priority_queue<ll, vector<ll>, greater<ll>> que1;
    priority_queue<ll> que2;
    for (int i = 0; i < 3 * N - 1; i++) {
        sum1[i + 1] = sum1[i];

        if (que1.size() < N) {
            que1.push(a[i]);
            sum1[i + 1] += a[i];
            continue;
        }

        if (que1.top() < a[i]) {
            sum1[i + 1] += a[i] - que1.top();
            que1.pop();
            que1.push(a[i]);
        }
    }

    for (int i = 3 * N - 1; i >= 0; i--) {
        sum2[i] = sum2[i + 1];

        if (que2.size() < N) {
            que2.push(a[i]);
            sum2[i] += a[i];
            continue;
        }

        if (que2.top() > a[i]) {
            sum2[i] += a[i] - que2.top();
            que2.pop();
            que2.push(a[i]);
        }
    }

    ll ans = -(1LL << 50);
    for (int i = N; i <= 2 * N; i++) {
        ans = max(ans, sum1[i] - sum2[i]);
    }
    cout << ans << endl;

    /*
    for (int i = 0; i < 3 * N; i++) {
        cerr << "[" << i << "] sum1:" << sum1[i] << " sum2:" << sum2[i] << endl;
    }
    */
    
    return 0;
}