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.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val a = (1..n).map { sc.nextInt() }
    val b = (1..n).map { sc.nextInt() }
    val permList = genPermList(n, listOf<Int>())
    println(Math.abs(permList.indexOf(a) - permList.indexOf(b)))
}

fun genPermList(n: Int, used: List<Int>): List<List<Int>> {
    val res = mutableListOf<List<Int>>()
    if (used.size == n) {
        res.add(used)
        return res
    }
    for (i in 1..n) {
        if (!used.contains(i)) {
            val nextUsed = mutableListOf<Int>()
            nextUsed.addAll(used)
            nextUsed.add(i)
            res.addAll(genPermList(n, nextUsed))
        }
    }
    return res
}