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() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    int a[330000];
    priority_queue<int, vector<int>, greater<int> > leftque;
    priority_queue<int> rightque;
    long long leftsum[330000], rightsum[330000], ans = -1e18;
    for (int i = 0; i < 3 * N; ++i) {
        cin >> a[i];
        if (i < N) {
            leftque.push(a[i]);
            leftsum[N - 1] += a[i];
        } else if (i >= 2 * N) {
            rightque.push(a[i]);
            rightsum[2 * N] += a[i];
        }
    }
    for (int i = N; i < 2 * N; ++i) {
        leftque.push(a[i]);
        int res = leftque.top();
        leftque.pop();
        leftsum[i] = leftsum[i - 1] - res + a[i];
    }
    for (int i = 2 * N - 1; i >= N; --i) {
        rightque.push(a[i]);
        int res = rightque.top();
        rightque.pop();
        rightsum[i] = rightsum[i + 1] - res + a[i];
    }
    for (int i = N; i < 2 * N; ++i) {
        ans = max(ans, leftsum[i - 1] - rightsum[i]);
    }
    cout << ans << endl;

    return 0;
}