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;

typedef long long int ll;

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n;
  cin >> n;
  vector < int > a(3 * n + 1);
  vector < ll > lt(3 * n + 1), rt(3 * n + 1);
  for(int i = 1; i <= 3 * n; i++) cin >> a[i];

  priority_queue < int, vector < int >, greater < int > > q;
  ll sum = 0;
  for(int i = 1; i < n; i++){
    q.push(a[i]);
    sum += a[i];
  }

  lt[n] = sum;
  for(int i = n + 1; i <= 2 * n; i++){
    q.push(a[i - 1]);
    sum += a[i - 1];
    sum -= q.top();
    q.pop();
    lt[i] = sum;
  }

  priority_queue < int > qq;
  sum = 0;
  for(int i = 2 * n + 1; i <= 3 * n; i++){
    qq.push(a[i]);
    sum += a[i];
  }

  rt[2 * n] = sum;
  for(int i = 2 * n - 1; i >= n; i--){
    q.push(a[i + 1]);
    sum += a[i + 1] - qq.top();
    qq.pop();
    rt[i] = sum;
  }

  ll ans = LLONG_MIN;
  for(int i = n; i <= 2 * n; i++){
    ans = max(ans, lt[i] + a[i] - rt[i]);
  }

  cout << ans << '\n';
}