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: Wrong Answer)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<bitset>
#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<tuple>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<stack>
#include<numeric>
#include<functional>
#include<iomanip>
#include<random>

#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define REP(i,n) for(int i = 0;i < n;i++)
#define FOR(i,a,b) for(int i = a;i < b;i++)
#define MEM(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ull, ull> UP;
typedef tuple<int, int, int> T;

int gcd(int a, int b) {
	if (a < b)swap(a, b);
	int k = a % b;
	if (k == 1)return 1;
	if (k == 0)return b;
	int ans = gcd(b, k);
	return ans;
}
int digisum(ll a) {
	int sum = 0;
	while (a) {
		sum += a % 10;
		a /= 10;
	}
	return sum;
}

int n;
int a[364364];
int main() {
	cin >> n;
	REP(i, 3 * n) {
		cin >> a[i];
	}
	ll sum0 = 0, sum1 = 0;
	priority_queue<int, vector<int>, greater<int>> que0;
	priority_queue<int> que1;
	REP(i, n) {
		que0.push(a[i]);
		sum0 += a[i];
	}
	FOR(i, 2 * n, 3 * n) {
		que1.push(a[i]);
		sum1 += a[i];
	}
	int last = n - 1;
	FOR(i, n, 2 * n) {
		int x = que0.top();
		if (x < a[i]) {
			last = i;
			sum0 += a[i] - x;
			que0.push(x);
			que0.pop();
		}
	}
	FOR(i, last + 1, 2 * n) {
		int y = que1.top();
		if (y > a[i]) {
			sum1 += a[i] - y;
			que1.push(y);
			que1.pop();
		}
	}
	cout << sum0 - sum1 << endl;
}