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: Accepted)

import java.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.io.PrintWriter
import java.lang.StringBuilder
import java.util.*

fun nextPermutation(array: MutableList<Int>): Boolean {
    var tmp = 0
    for (i in array.size-2 downTo 0) {
        if (array[i] < array[i+1]) {
            for (j in array.size-1 downTo i+1) {
                if (array[j] > array[i]) {
                    tmp = array[i]
                    array[i] = array[j]
                    array[j] = tmp

                    val tmpArray = array.takeLast(array.size-(i+1)).sorted()
                    tmpArray.forEachIndexed { index, e ->
                        array[i+1+index] = e
                    }

                    return true
                }
            }
        }
    }

    return false
}

fun PrintWriter.solve(sc: FastScanner) {
    val n = sc.nextInt()
    val p = Array(n) { sc.nextInt() }
    val q = Array(n) { sc.nextInt() }
    val a = (1..n).toMutableList()
    var ans = 0
    do {
        if ((0 until n).all { p[it] == a[it] }) {
            break
        }
        ans++
    } while (nextPermutation(a))
    a.sort()
    do {
        if ((0 until n).all { q[it] == a[it] }) {
            break
        }
        ans--
    } while (nextPermutation(a))
    println(Math.abs(ans))
}

fun main(args: Array<String>) {
    val writer = PrintWriter(System.out, false)
    writer.solve(FastScanner(System.`in`))
    writer.flush()
}

class FastScanner(s: InputStream) {
    private var st = StringTokenizer("")
    private val br = BufferedReader(InputStreamReader(s))

    fun next(): String {
        while (!st.hasMoreTokens()) st = StringTokenizer(br.readLine())

        return st.nextToken()
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextLine() = br.readLine()
    fun nextDouble() = next().toDouble()
    fun ready() = br.ready()
}