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: Wrong Answer)

#include<stdio.h>
void push(long int x[],long int i,long int k){//最小が根になる x: 配列，i:push後のデータ数，k: pushするデータ
  x[i]=k;
  for(int j=i;j>1;j/=2)if(x[j]<x[j/2])x[j]^=x[j/2]^=x[j]^=x[j/2];
}
long int pop(long int x[],long int i){//最初値を取り出す x: 配列，i:pop前のデータ数
  int j=1,k;
  x[0]=x[1];x[1]=x[i];
  while(j*2<i&&(x[j*2]<x[j]||x[j*2+1]<x[j])){
    k=x[j*2]<x[j*2+1]?j*2:j*2+1;
    x[j]^=x[k]^=x[j]^=x[k];
    j=k;
  }
  return x[0];
}
int main(){
  long int a[300000],x[100001],y[100001],k[100001],l[100001];
  long int i,n,j;
  scanf("%ld",&n);
  for(i=0;i<n*3;i++)scanf("%ld",&a[i]);
  k[0]=0;
  for(i=0;i<n;i++){push(x,i+1,a[i]);k[0]+=a[i];}
  for(i=0;i<n;i++){k[i+1]=k[i]+a[n+i]-pop(x,n);push(x,n,a[n+i]);}
  k[n]=0;
  for(i=0;i<n;i++){push(y,i+1,-a[n*3-i-1]);l[n]-=a[n*3-i-1];}
  for(i=0;i<n;i++){l[n-i-1]=l[n-i]-a[n*2-i-1]-pop(y,n);;push(y,n,a[n*2-i-1]);}
  j=k[0]+l[0];
  for(i=0;i<n;i++)j=j<k[i+1]+l[i+1]?k[i+1]+l[i+1]:j;
  printf("%ld",j);
}