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)

void solve();

int main() {
	solve();
	return 0;
}

//////////////////////////////////////////////////
//////////////////////////////////////////////////
#include <iostream>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;
int N;
int A[300000];
long long sum1[100001];   //0 - N-1,0 - N,0 - N+1 ,...からN個取ったときの和の最大値
long long sum2[100001];   //2*N - 3*N-1,2*N-1 - 3*N-1,...,2*N-i - 3*N-1,...の和の最小値
priority_queue<int, vector<int>, greater<int>> q1;
priority_queue<int> q2;

void solve() {
	cin >> N;
	for (int n = 0; n < 3 * N; n++)cin >> A[n];
	sum1[0] = 0;
	for (int n = 0; n < N; n++) {
		sum1[0] += A[n];
		q1.push(A[n]);
	}
	for (int n = N; n < 2 * N; n++) {
		if (q1.top() < A[n]) {
			sum1[n-N+1] = sum1[n-N]+A[n]-q1.top();
			q1.pop();
			q1.push(A[n]);
		}else sum1[n - N + 1] = sum1[n - N] ;
	}
	sum2[0] = 0;
	for (int n = 3 * N - 1; n >= 2 * N; n--) {
		sum2[0] += A[n];
		q2.push(A[n]);
	}
	for (int i = 1; i <= N; i++) {
		if (q2.top() > A[2*N-i]) {
			sum2[i] = sum2[i-1] + A[2 * N - i] - q2.top();
			q2.pop();
			q2.push(A[2 * N - i]);
		}
		else sum2[i] = sum2[i-1];
	}
	long long ans = -100000000000000;
	for (int n = 0; n <= N; n++) {
		if (ans < sum1[n] - sum2[N - n])ans = sum1[n] - sum2[N - n];
	}
	cout << ans << endl;
	return;
}