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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <math.h>
#include <stack>
#include <queue>
#include <functional>
#include <limits.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define rep(i,N) for (int i=0;i<N;i++)

//変数
int N;
vector<int> a;

//解法
void solve() {

	ll ans = LLONG_MIN;
	rep(i, N + 1) {
		vector<int> front, back;
		rep(j, N*3) {
			if (j < N + i) {
				front.emplace_back(a[j]);
			}
			else {
				back.emplace_back(a[j]);
			}
		}

		sort(front.begin(), front.end(), less<int>());
		sort(back.begin(), back.end(), greater<int>());

		front.erase(front.begin(), front.begin() + i);
		back.erase(back.begin(), back.begin() + (N - i));

		int size = front.size();
		ll sum1=0, sum2=0,sum;
		rep(j, size) sum1 += front[j];
		size = back.size();
		rep(j, size) sum2 += back[j];

		sum = sum1 - sum2;
		if (ans < sum) ans = sum;
	}

	cout << ans;

	return;
}

int main() {
	//入力
	cin >> N;
	a.resize(3*N);
	rep(i, 3*N) cin >> a[i];
	solve();
	cout << endl;
	return 0;
}