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 Scala (2.11.7) (judge verdict: Wrong Answer)

import java.util.Scanner
import scala.collection.mutable.PriorityQueue

object Main extends App {
  val sc = new Scanner(System.in)

  val n = sc.nextInt()
  val a = new Array[Int](3*n + 1)

  for(i <- 1 to 3*n)
    a(i) = sc.nextInt()

  val left = new Array[Long](n+1)
  val right = new Array[Long](n+1)


  // left
  left(0) = a.slice(1, n+1).sum
  val pqd = new PriorityQueue[Int]()(Ordering.Int.reverse)
  for(i <- 1 to n)
    pqd.enqueue(a(i))
  for(i <- n+1 to 2*n){
    pqd.enqueue(a(i))
    left(i-n) = left(i-n-1) + a(i) - pqd.dequeue
  }


  // right
  right(n) = a.slice(2*n+1, 3*n+1).sum
  val pqa = new PriorityQueue[Int]
  for(i <- (2*n+1 to 3*n).reverse)
    pqa.enqueue(a(i))
  for(i <- (n to 2*n-1).reverse){
    pqa.enqueue(a(i+1))
    right(i-n) = right(i-n+1) + a(i+1) - pqa.dequeue
  }


  var ans = -1e18.toLong
  for(i <- 0 to n){
    ans = Math.max(ans, left(i) - right(i))
  }

  println(ans)
}