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;  
#define INF 0x3f3f3f3f  
typedef long long ll;  
const int N = 1e5 + 10;  
ll Lft[3*N], Rit[3*N];
int num[3*N];
int main()  
{  
  //  freopen("in.txt", "r", stdin);
   	int n;
   	while (~scanf("%d", &n))
   	{
   		for (int i = 0; i < 3*n; i++)
   			scanf("%d", num + i);
   		priority_queue<int, vector<int>, greater<int> >qmin; 
   		priority_queue<int> qmax;
   		ll sum = 0;
   		qmin.push(num[0]);

   		for (int i = 0; i < n; i++)
   		{
   			sum += num[i];
   			qmin.push(num[i]);
   		}
   		int m = n<<1;
   		Lft[n-1] = sum;
   		for (int i = n; i < m; i++)
   		{
   			if (qmin.top() < num[i])
   			{
   				sum += num[i];
   				sum -= qmin.top();
   				qmin.pop();
   				qmin.push(num[i]);
   			}
   			Lft[i] = sum;
   		}

   		sum = 0;
   		for (int i = 3*n - 1; i >= m; i--)
   		{
   			sum += num[i];
   			qmax.push(num[i]);
   		}
   		Rit[m] = sum;
   		for (int i = m - 1; i >= n; i--)
   		{
   			if (num[i] < qmax.top())
   			{
   				sum += num[i];
   				sum -= qmax.top();
   				qmax.pop();
   				qmax.push(num[i]);
   			}	
   			Rit[i] = sum;
   		}
   		ll ans = Lft[n-1] - Rit[n];
   		for (int i = n; i < m; i++)
   		{
   			// if (n == 1)
   			// printf("%d %d\n", Lft[i-1], Rit[i]);
   			ans = max(Lft[i-1] - Rit[i], ans);
   		}
   		printf("%lld\n", ans);
   	}

   return 0;  
}