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)

#pragma region template
#include "bits/stdc++.h"
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
const long long MOD = 1e9 + 7, MAX = 1e18, larg = 1e5, INF = -1e9;
long long A, B, C, D, E, F, G, H, I, J, K, L, N, M, O, P, Q, R, S, T, U, V, W, x, y, z;
long long max_value = INF, max_index = -1;
long long min_value = MAX, min_index = -1;
typedef long long ll;
typedef pair<ll, ll> Pa;

ll arr[300000];
ll val[100001];
int main() {
	cin >> N;
	rep(i, 3 * N) {
		cin >> arr[i];
	}

	priority_queue<ll, vector<ll>, greater<ll>> que_a;
	ll sum = 0;
	rep(i, N) {
		que_a.push(arr[i]);
		sum += arr[i];
	}
	val[0] = sum;
	rep(i, N) {
		sum += arr[N + i];
		que_a.push(arr[N + i]);
		sum -= que_a.top();
		que_a.pop();
		val[i + 1] = sum;
	}

	priority_queue<ll> que_b;
	sum = 0;
	rep(i, N) {
		que_b.push(arr[2 * N + i]);
		sum += arr[2 * N + i];
	}
	ll maxv = val[N] - sum;
	for (int i = N - 1; i >= 0; i--) {
		sum += arr[N + i];
		que_b.push(arr[N + i]);
		sum -= que_b.top();
		que_b.pop();
		maxv = max(maxv, val[i] - sum);
	}
	cout << maxv;
}