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>
#include<algorithm>

using namespace std;

typedef long long ll;

ll N, a[100000 * 3 + 10];
ll bleft[100000 + 1 + 10], bright[100000 + 1 + 10];

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

	priority_queue <ll, vector <ll>, greater<ll>> pq1;
	for (int i = 0; i < N; i++)pq1.push(a[i]), bleft[0] += a[i];
	for (int i = N; i < 2 * N; i++) {
		bleft[i + 1 - N] = max(bleft[i - N], bleft[i - N] - pq1.top() + a[i]);
		if (pq1.top() < a[i])pq1.pop(), pq1.push(a[i]);
	}

	priority_queue<ll> pq2;
	for (int i = 2 * N; i < 3 * N; i++)pq2.push(a[i]), bright[N] += a[i];
	for (int i = 2 * N - 1; i >= N; i--) {
		bright[i - N] = min(bright[i + 1 - N], bright[i + 1 - N] - pq2.top() + a[i]);
		if (pq2.top() > a[i])pq2.pop(), pq2.push(a[i]);
	}
	ll ans = -114514919810893;
	for (int i = 0; i <= N; i++) {
		ans = max(ans, bleft[i] - bright[i]);
	}
	cout << ans << endl;
	return 0;
}