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: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
public class Main
{
	public static void main(String args[])throws IOException {
		BufferedReader br  =new BufferedReader(new InputStreamReader(System.in));
		int n =Integer.parseInt(br.readLine());
		long a[]=new long[0];
		a=Arrays.stream(br.readLine().split("\\s+")).mapToLong(Long::parseLong).toArray();
	    PriorityQueue<Long> npq = new PriorityQueue<>(Comparator.naturalOrder());
	    PriorityQueue<Long> rpq = new PriorityQueue<>(Comparator.reverseOrder());
	    long[] s = new long[n+1];
	    long[] t = new long[n+1];
	    long su = 0L;
	    long tu = 0L;
	    for (int i=0;i<n;i++) {
	      npq.offer(a[i]);
	      su += a[i];
	      rpq.offer(a[3*n-1-i]);
	      tu += a[3*n-1-i];
	    }
	    s[0] = su;
	    t[n] = tu;
	    for (int i=n;i<2*n;i++) {
	      npq.offer(a[i]);
	      su += a[i]-npq.poll();
	      s[i-n+1] = su;
	      rpq.offer(a[3*n-1-i]);
	      tu += a[3*n-1-i]-rpq.poll();
	      t[2*n-i-1] = tu;
	    }
	    long m = s[0]-t[0];
	    for (int i=0;i<n+1;i++) {
	      if (m < s[i]-t[i]) m = s[i]-t[i];
	    }
	    System.out.println(m);
	
	}
}