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>
#define LL long long
using namespace std;
const int maxn=1e5+10;
LL L[3*maxn];
LL R[3*maxn];
int a[3*maxn];
int n;
int main()
{
	priority_queue<int,vector<int>,greater<int> > Q;
	scanf("%d",&n);
	LL sum=0;
	for(int i=1;i<=3*n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
	{
		sum+=(LL)a[i];
		Q.push(a[i]);
		L[i]=sum;
	}
	for(int i=n+1;i<=2*n;i++)
	{
		if(a[i]>Q.top())
		{
			sum-=Q.top();
			Q.pop();
			sum+=a[i];
			Q.push(a[i]);
		}
		L[i]=sum;
	}

	sum=0;
	while(!Q.empty()) Q.pop();
	for(int i=3*n;i>2*n;i--)
	{
		sum+=(LL)a[i];
		Q.push(a[i]);
		R[i]=sum;
	}
	for(int i=2*n;i>n;i--)
	{
		if(a[i]>Q.top())
		{
			sum-=Q.top();
			Q.pop();
			sum+=a[i];
			Q.push(a[i]);
		}
		R[i]=sum;
	}
	LL ans=0;
	for(int i=n;i<=2*n;i++)
		ans=max(ans,L[i]+R[i+1]);
	printf("%lld",ans);
}