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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <sstream>
#include <iomanip> 
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <math.h> 

using namespace std;



int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	long long n, i, j, k, ans, n3, s, x;
	cin >> n;
	n3 = 3 * n;
	vector<long long> a(n3), ll(n3, 0), rr(n3, 0);
	for (i = 0; i < n3; i++) cin >> a[i];

	priority_queue<long long> pq;

	s = 0;
	for (i = 0; i < n; i++) {
		pq.push(-a[i]);
		s += a[i];
	}
	ll[n - 1] = s;

	for (i = n; i < n * 2; i++) {
		pq.push(-a[i]);
		s += a[i];

		x = -pq.top();
		s -= x;
		pq.pop();

		ll[i] = s;
	}


	while (!pq.empty()) pq.pop();

	s = 0;
	for (i = n3 - 1; i >= n * 2; i--) {
		pq.push(a[i]);
		s += a[i];
	}
	rr[n * 2] = s;

	for (i = n * 2 - 1; i >= n; i--){
		pq.push(a[i]);
		s += a[i];

		x = pq.top();
		s -= x;
		pq.pop();

		rr[i] = s;
	}

	ans = -(1LL << 61);
	for (i = n - 1; i < n * 2; i++) ans = max(ans, ll[i] - rr[i + 1]);

	cout << ans << "\n";


	return 0;
}