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 rep(i, n) for(int i = 0; i < (n); ++i)
#define repr(i, n) for(int i = (n-1); i >= 0; --i)

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 200005;
const int INF = 1001001001;
const int MOD = 1000000007;

int main(){
    int N;
    cin >> N;

    vector<int> A(3*N);
    rep(i, 3*N) cin >> A[i];

    priority_queue<int, vector<int>, greater<int>> L;
    vector<ll> sumL(3*N+1);
    sumL[0] = 0;
    rep(i, 3*N) {
        ll res = 0;
        L.push(A[i]);
        if (L.size() <= N) {
            res = sumL[i] + A[i];
        } else {
            res = sumL[i] + A[i] - L.top();
            L.pop();
        }
        sumL[i + 1] = res;
    }

    priority_queue<int> R;
    vector<ll> sumR(3*N+1);
    sumR[3*N] = 0;
    repr(i, 3*N) {
        ll res = 0;
        R.push(A[i]);
        if (R.size() <= N) {
            res = sumR[i+1] + A[i];
        } else {
            res = sumR[i+1] + A[i] - R.top();
            R.pop();
        }
        sumR[i] = res;        
    }

    ll ans = -(ll)INF * INF;
    for (int i = N; i <= 2*N; ++i) {
        ans = max(ans, sumL[i] - sumR[i]);
    }

    cout << ans << endl;
}