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: Wrong Answer)

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val P = readLine()!!.split(" ").map(String::toInt).toMutableList()
    val Q = readLine()!!.split(" ").map(String::toInt).toMutableList()
    val score1 = calc(P, N)
//    val score2 = calc(Q, N)
//    println(Math.abs(score1 - score2))
}

fun calc(a: MutableList<Int>, N: Int): Int {
    var score = 1
//    for (i in 0 until N) {
//        score += (a[i] - 1) * fact(N - i - 1)
//        for (j in i+1 until N) {
//            if (a[i] < a[j]) a[j]--
//        }
//    }
    return score
}