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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main(){
  long n, a, sumf = 0, sumr = 0;  cin >> n;
  priority_queue<long, vector<long>, greater<long>> fr;
  vector<int> mid(n);
  priority_queue<long> re;  

  for(int i=0; i<n; i++){
    cin >> a;
    sumf += a;
    fr.push(a);
  }
  for(int i=0; i<n; i++)  cin >> mid[i];
  for(int i=0; i<n; i++){
    cin >> a;
    sumr += a;
    re.push(a);
  }
  vector<long> sf(n+1), sr(n+1);  sf[0] = sumf, sr[n] = sumr;
  for(int i=0; i<n; i++){
    sumf -= fr.top();  fr.pop();
    sumf += mid[i];
    fr.push(mid[i]);
    sf[i+1] = sumf;
  }
  for(int i=n-1; i>=0; i--){
    sumr -= re.top();  re.pop();
    sumr += mid[i];
    re.push(mid[i]);
    sr[i] = sumr;
  }
  long ans = -1e18;
  for(int i=0; i<=n; i++)
    ans = max(ans, sf[i]-sr[i]);
    
  cout << ans << endl;

  return 0;
}