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 Kotlin (1.0.0) (judge verdict: Runtime Error)

import java.util.*

fun main() {
  val sc = Scanner(System.`in`)
  val n = sc.nextInt()
  val p = (1..n).map { sc.nextInt() }
  val q = (1..n).map { sc.nextInt() }

  println(abc150c(n, p, q))
}

private fun abc150c(n: Int, p: List<Int>, q: List<Int>): Int {
  val pStr = p.joinToString("")
  val qStr = q.joinToString("")
  val array = (1..n).map { it }.toIntArray()

  var count = 1
  var a = 0
  var b = 0
  do {
    val arrayStr = array.joinToString("")
    if (arrayStr == pStr) a = count
    if (arrayStr == qStr) b = count

    count++
  } while (findNextPermutation(array))

  return Math.abs(a - b)
}

private fun findNextPermutation(array: IntArray): Boolean {
  if (array.size <= 1) return false

  var last = array.size - 2
  while (last >= 0) {
    if (array[last] < array[last + 1]) break
    last--
  }
  if (last < 0) return false

  var nextGreater = array.size - 1
  var i = array.size - 1
  while (i > last) {
    if (array[i] > array[last]) {
      nextGreater = i
      break
    }
    i--
  }
  swap(array, nextGreater, last)
  reverse(array, last + 1, array.size - 1)

  return true
}

private fun swap(array: IntArray, left: Int, right: Int): IntArray {
  val tmp = array[left]
  array[left] = array[right]
  array[right] = tmp
  return array
}

private fun reverse(array: IntArray, left: Int, right: Int): IntArray {
  var l = left
  var r = right
  while (l < r) {
    val tmp = array[l]
    array[l++] = array[r]
    array[r--] = tmp
  }
  return array
}