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: Time Limit Exceeded)

#include<vector>
#include<list>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long

int main(){
  ll N, a;
  cin >> N;
  vector<ll> A;
  for(int i=0; i<3*N; ++i) {
    cin >> a;
    A.emplace_back(a);
  }

  list<ll> min(A.begin(), A.begin()+N), max(A.end()-N, A.end());

  min.sort();
  max.sort();

  vector<ll> S_l{0}, S_r{0};
  for(int i=0; i<N; ++i){
    S_l[0]+=A[i];
    S_r[0]+=A[2*N+i];
  }

  ll s_l, s_r;
  auto ptr_min = min.begin();
  auto ptr_max = --max.end();
  for(int i=0; i<N; ++i) {
    a = A[N+i];
    if (a <= *ptr_min){
      s_l = a;
    }else{
      s_l = *ptr_min;
      auto ptr_insert = lower_bound(ptr_min, min.end(), a);
      min.insert(ptr_insert, a);
      ++ptr_min;
    }
    S_l.emplace_back(S_l[i] + a - s_l);

    a = A[2*N-i-1];
    if (a >= *ptr_max){
      s_r = a;
    }else{
      s_r = *ptr_max;
      auto ptr_insert = upper_bound(max.begin(), ptr_max, a);
      max.insert(ptr_insert, a);
      --ptr_max;
    }
    S_r.emplace_back(S_r[i] + a - s_r);
  }

  ll ans = S_l[0]-S_r[N];
  for(int i=1; i<N+1; ++i) {
    if(S_l[i]-S_r[N-i] > ans) ans = S_l[i]-S_r[N-i];
  }

  cout << ans << endl;
  
  return 0;
}