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 <sstream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
#include <utility>
#include <set>
#include <list>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <iterator> 
#include <limits>

#define REP(i,n) for (int i=0;i<(n);i++)
#define FOR(i,a,b) for (int i=(a); i<(b); i++)
#define ALL(a) (a).begin(),(a).end()
#define RALL(A) (a).rbegin(),(a).rend()
#define PB push_back
#define MP make_pair

#define dump(x) cerr << #x << " = " << (x) << endl;

using namespace std;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = numeric_limits<int>::max() / 2;
const int NEG_INF = numeric_limits<int>::min() / 2;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

int N;
VI A;

int main(int argc, char const* argv[])
{
    cin >> N;
    REP(i, 3 * N) {
        int a; cin >> a; A.PB(a);
    }
    map<int, LL> left_sum_save;
    LL left_sum = 0;
    priority_queue<int> pq;
    REP(i, N) {
        pq.push(-A[i]);
        left_sum += A[i];
    }
    left_sum_save[0] = left_sum;
    REP(i, N) {
        pq.push(-A[i + N]);
        int top = pq.top();
        pq.pop();
        top = -top;
        left_sum = left_sum + A[i + N] - top;
        left_sum_save[i + 1] = left_sum;
    }

    map<int, LL> right_sum_save;
    LL right_sum = 0;
    priority_queue<int> pq2;
    REP(i, N) {
        pq2.push(A[i + 2 * N]);
        right_sum += A[i + 2 * N];
    }
    right_sum_save[0] = right_sum;
    REP(i, N) {
        pq2.push(A[2 * N - i - 1]);
        int top = pq2.top(); pq2.pop();
        right_sum = right_sum + A[2 * N - i - 1] - top;
        right_sum_save[i + 1] = right_sum;
    }

    LL ans = NEG_INF;
    REP(i, N) {
        LL temp = left_sum_save[i] - right_sum_save[N - i];
        ans = max(ans, temp);
    }
    cout << ans << endl;

    return 0;
}