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.lang.Math

val MOD: Int = 1000000000 + 7
val MODL: Long = MOD.toLong()

var fact = mutableListOf<Int>()

fun main(args: Array<String>)
{
    val n = read().toInt()
    calcFact(n)
    val p = read().split(" ").map{it.toInt()}.toIntArray()
    val q = read().split(" ").map{it.toInt()}.toIntArray()

    var a = 1
    for (i in 0 until n) {
        a += (p[i] - 1) * fact[n-1-i]
        for (k in i+1 until n) {
            if (p[k] > p[i]) p[k] -= 1
        }
    }

    var b = 1
    for (i in 0 until n) {
        b += (q[i] - 1) * fact[n-1-i]
        for (k in i+1 until n) {
            if (q[k] > q[i]) q[k] -= 1
        }
    }

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

fun calcFact(n: Int): Int {
    if (n == 0) {
        fact.add(1)
        return 1
    }

    val f = n * calcFact(n-1)
    fact.add(f)
    return f
}

fun read(): String = readLine()!!