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)


#include "bits/stdc++.h"
using namespace std;

#define DEBUG(x) cout<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cout<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl


typedef long long ll;
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rrep1(i,n) for(int i=n;i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(v) v.begin(),v.end()
#define UNIQUE(v) v.erase(unique(all(v)), v.end())
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
const ll inf = 1000000001;
const ll INF = 1e18;
const ll MOD = 1000000007;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { 1,0, -1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
//#define mp make_pair


int main() {
	int n;
	cin >> n;
	vl a(3 * n);
	rep(i, 3 * n) {
		cin >> a[i];
	}
	priority_queue<ll> pq;
	vl leftsum(3 * n, -INF), rightsum(3 * n, INF);
	leftsum[n - 1] = 0; rightsum[2 * n - 1] = 0;
	rep(i, n) {
		leftsum[n - 1] += a[i];
		pq.push(-a[i]);
	}
	REP(i, n, 2 * n) {
		leftsum[i] = leftsum[i - 1];
		leftsum[i] += a[i];
		pq.push(-a[i]);
		leftsum[i] += pq.top();
		pq.pop();
	}
	priority_queue<ll> pqr;
	for (int i = 3 * n - 1; i >= 2 * n; i--) {
		rightsum[2 * n - 1] += a[i];
		pqr.push(a[i]);
	}	
	for (int i = 2 * n - 1; i >= n; i--) {
		rightsum[i - 1] = rightsum[i];
		rightsum[i - 1] += a[i];
		pqr.push(a[i]);
		rightsum[i - 1] -= pqr.top();
		pqr.pop();
	}
	ll ans = -INF;
	REP (i, n - 1, 2 * n) {
		chmax(ans, leftsum[i] - rightsum[i]);
	}
	cout << ans << endl;
}