Problem:
Score : 300 points

Problem Statement

We have two permutations P and Q of size N (that is, P and Q are both rearrangements of (1,~2,~...,~N)).

There are N! possible permutations of size N. Among them, let P and Q be the a-th and b-th lexicographically smallest permutations, respectively. Find |a - b|.

Notes

For two sequences X and Y, X is said to be lexicographically smaller than Y if and only if there exists an integer k such that X_i = Y_i~(1 \leq i < k) and X_k < Y_k.

Constraints

2 \leq N \leq 8

P and Q are permutations of size N.

Input

Input is given from Standard Input in the following format:

N
P_1 P_2 ... P_N
Q_1 Q_2 ... Q_N

Output

Print |a - b|.

Sample Input 1

3
1 3 2
3 1 2

Sample Output 1

3

There are 6 permutations of size 3: (1,~2,~3), (1,~3,~2), (2,~1,~3), (2,~3,~1), (3,~1,~2), and (3,~2,~1). Among them, (1,~3,~2) and (3,~1,~2) come 2-nd and 5-th in lexicographical order, so the answer is |2 - 5| = 3.

Sample Input 2

8
7 3 5 4 2 1 6 8
3 8 2 5 4 6 7 1

Sample Output 2

17517

Sample Input 3

3
1 2 3
1 2 3

Sample Output 3

0

Solution in Scala (2.11.7) (judge verdict: Runtime Error)

import scala.collection.mutable.ListBuffer
import scala.io.StdIn

object Main extends App {

  val l = StdIn.readLine()
  val s1 = StdIn.readLine().split(" ").toString
  val s2 = StdIn.readLine().split(" ").toString

  println(math.abs(impl(s1) - impl(s2)))

  def impl(s: String): Int = {
    val fact0 = 0
    val fact1 = 1
    val fact2 = 1 * 2
    val fact3 = 1 * 2 * 3
    val fact4 = 1 * 2 * 3 * 4
    val fact5 = 1 * 2 * 3 * 4 * 5
    val fact6 = 1 * 2 * 3 * 4 * 5 * 6
    val fact7 = 1 * 2 * 3 * 4 * 5 * 6 * 7
    val fact8 = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8

    val fact = Array(fact0, fact1, fact2, fact3, fact4, fact5, fact6, fact7, fact8)

    var pendingLen = s.length - 1
    var lowestPending = 1
    val seenSoFar = new ListBuffer[Int]()

    var order = 0

    for (c <- s) {
      val num = c.toString.toInt

      if(num != lowestPending) {
        order += ((num - lowestPending - seenSoFar.count(_ < num)) * fact(pendingLen))
      } else {
        lowestPending += 1
        while (seenSoFar.contains(lowestPending)) {
          lowestPending += 1
        }
      }

      seenSoFar.append(num)
      pendingLen -= 1
    }

    order
  }
}