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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MAX(a,b) (((a)>(b))?(a):(b))
#define ll long long

int heap[100001];
int num=0;
void push(int a){
 	int i=num++;
 	while(i>0){
 		int p=(i-1)/2;
 		if(heap[p]<=a) break;
 		heap[i]=heap[p];
 		i=p;
 	}
 	heap[i]=a;
 	return;
}
int pop(void){
 	int data=heap[0];
 	int n=heap[--num];
 	int pos=0;
 	while(pos*2+1<num){
 		int a=pos*2+1;
 		int b=pos*2+2;
 		if(b<num&&heap[a]>heap[b]) a=b;
 		if(n<=heap[a]) break;
 		heap[pos]=heap[a];
 		pos=a;
 	}
 	heap[pos]=n;
 	return data;
}
int a[300000];
ll k[100001];
int main(void)
{
     int n,p,size,i;
     ll sum=0,res;
     scanf("%d",&n);
     size=3*n;
     FOR(i,0,size) scanf("%d",&a[i]);
     FOR(i,0,n){
     	sum+=a[i];
     	push(a[i]);
     }
     k[0]=sum;
     FOR(i,1,n+1){
     	push(a[i+n-1]);
     	p=pop();
     	sum+=(a[i+n-1]-p);
     	k[i]=sum;
     }
     sum=num=0;
     FOR(i,0,n){
     	sum+=a[size-i-1];
     	push(-a[size-i-1]);
     }
     res=k[n]-sum;
     FOR(i,1,n+1){
     	push(-a[2*n-i]);
     	p=-pop();
     	sum+=(a[2*n-i]-p);
     	res=MAX(res,k[n-i]-sum);
     }
     printf("%lld\n",res);
 	return 0;
}