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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

// 昇順
int compa(const void *a, const void *b) { return *(int *)a - *(int *)b; }
// 降順
int compd(const void *a, const void *b) { return *(int *)b - *(int *)a; }

int N;
int a[300000];
int left[100000];
int right[100000];

int main(void) {
	int i, left_i = 0, right_i = 0, left_ii = 0, right_ii = 0;
	long long int ans = 0;
	scanf("%d", &N);
	for (i = 0; i < N * 3; i++) {
		scanf("%d", &a[i]);
		if (i < N) {
			left[i] = a[i];
		}
		else if (i >= 2 * N) {
			right[i - 2 * N] = a[i];
		}
	}

	for (i = 0; i < N; i++) {
		ans += (a[i] - a[3 * N - 1 - i]);
	}
	int left_v, right_v;
	qsort(left, N, sizeof(int), compa);
	qsort(right, N, sizeof(int), compd);
	for (i = 0; i < N; i++) {
		left_v = a[left_i + left_ii + N] - left[left_i];
		right_v = right[right_i] - a[2 * N - 1 - right_i - right_ii];
		if (left_v < 0 && right_v < 0) {
			if (left_v > right_v) {
				left_ii++;
			}
			else {
				right_ii++;
			}
		}
		else if (left_v > right_v) {
			left_i++;
			ans += left_v;
		}
		else {
			right_i++;
			ans += right_v;
		}
	}

	printf("%lld\n", ans);

	return 0;
}