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 <bits/stdc++.h>

using namespace std;
using ll = long long;
using pii = pair<int, int>;

int N;
ll a[300010];

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  cin >> N;
  for (int i = 0; i < 3 * N; i++) {
    cin >> a[i];
  }

  ll L[100010], R[100010];
  priority_queue<ll, vector<ll>, greater<ll>> queL;
  priority_queue<ll> queR;
  ll sumL = 0, sumR = 0;
  for (int i = 0; i < N; i++) {
    sumL += a[i];
    queL.emplace(a[i]);
  }
  L[0] = sumL;
  for (int i = 2 * N; i < 3 * N; i++) {
    sumR += a[i];
    queR.emplace(a[i]);
  }
  R[0] = sumR;
  for (int i = N; i < 2 * N; i++) {
    queL.emplace(a[i]);
    sumL += a[i];
    ll p = queL.top(); queL.pop();
    sumL -= p;
    L[i - N + 1] = sumL;
  }
  for (int i = 2 * N - 1; i >= N; i--) {
    queL.emplace(a[i]);
    sumR += a[i];
    ll p = queR.top(); queR.pop();
    sumR -= p;
    R[2 * N - i] = sumR;
  }

  ll ans = -(1LL << 60);
  for (int i = 0; i <= N; i++) {
    ans = max(ans, L[i] - R[N - i]);
  }
  cout << ans << endl;

  return 0;
}