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)

var permList = mutableListOf<String>()

fun main(args: Array<String>) {

    val N = readLine()!!.toInt()
    val PN = readLine()!!.split(" ").joinToString("")
    val QN = readLine()!!.split(" ").joinToString("")

    perm(N)

    val a = permList.indexOf(PN) + 1
    val b = permList.indexOf(QN) + 1

    println(Math.abs(a - b))
}

fun perm(n: Int){
    val c = (1..n).toList()
    val p = emptyList<Int>()
    perm(c, p)
}

fun perm(c: List<Int>, p: List<Int>) {
    if (c.isEmpty()) {
        permList.add(p.joinToString(""))
        return
    }
    for (e in c) {

        val newC = c.minusElement(e)
        val newP = p.plusElement(e)

        perm(newC, newP)
    }
}