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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
 
public class Main {
    static Scanner scanner = new Scanner(System.in);
 
    public static void main(String[]$) {
        int n = scanner.nextInt();
        int m = n * 3;
        long[] a = new long[m];
        for (int i = 0; i < m; i++) {
            a[i] = scanner.nextInt();
        }
        PriorityQueue<Long> queue = new PriorityQueue<>();
        long[] max = new long[n + 1];
        long sum = 0;
        for (int i = 0; i < m; i++) {
            queue.add(a[i]);
            if (i >= n && i <= n * 2) {
                max[i - n] = sum;
                sum -= queue.poll();
            }
            sum += a[i];
        }
 
        queue = new PriorityQueue<>(Comparator.reverseOrder());
        long[] min = new long[n + 1];
        sum = 0;
        for (int i = m; i > 0; i--) {
            queue.add(a[i - 1]);
            if (i >= n && i <= n * 2) {
                min[i - n] = sum;
                sum -= queue.poll();
            }
            sum += a[i - 1];
        }
        long ans = Integer.MIN_VALUE;
        for (int i = 0; i <= n; i++) {
            ans = Math.max(ans, max[i] - min[i]);
        }
        System.out.println(ans);
    }
}