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 <iostream>
#include <vector>
#include <queue>
using namespace std;

typedef long long ll;

ll solve(const int n, const vector<int> & in){
  ll lsum = 0, rsum = 0;
  priority_queue<int, vector<int>, greater<int> > lq;
  priority_queue<int> rq;

  for(int i = 0;i < n;i++){
    lq.push(in[i]);
    lsum += in[i];
  }
  for(int i = 2 * n;i < 3*n;i++){
    rq.push(in[i]);
    rsum += in[i];
  }
  ll ans = lsum - rsum;
  vector<ll> forward_sum(3*n, 0);
  vector<ll> backward_sum(3*n, 0);

  forward_sum[n-1] = lsum;
  for(int i = n;i < 2*n;i++){
    if (lq.top() < in[i]){
      lsum = lsum - lq.top() + in[i];
      lq.pop();
      lq.push(in[i]);
    }
    forward_sum[i] = lsum;
  }

  backward_sum[2*n] = rsum;
  for(int i = 2*n-1;i >= n;i--){
    if (rq.top() > in[i]){
      rsum = rsum - rq.top() + in[i];
      rq.pop();
      rq.push(in[i]);
    }
    backward_sum[i] = rsum;
  }


  for(int i = n-1;i < 2*n;i++){
    ans = max(ans, forward_sum[i] - backward_sum[i+1]);
  }
  return ans;
}


int main(){
  int n;
  cin >> n;
  vector<int> in(3*n);
  for(int i = 0;i <3* n;i++){
    cin >> in[i];
  }
  cout << solve(n, in) << endl;
  return 0;
}