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: Time Limit Exceeded)

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val p = readLine()!!.replace(" ", "").toInt()
    val q = readLine()!!.replace(" ", "").toInt()

    var pPos = 0
    var qPos = 0
    fun chk(i:Int) : Boolean{
        val s = i.toString().toCharArray()
        val f = BooleanArray(N + 1){ false }
        s.forEach {
            val c = it.toString().toInt()
            when {
                f[c] -> return false
                else -> f[c] = true
            }
        }
        return true
    }

    val a = IntArray(N)
    var cnt = 0
    fun create(n:Int) {
        for (m in 1 .. N) {
            a[n]= m
            if (n+1 == N) {
                val i = a.joinToString ("" ).toInt()
                if (chk(i)) {
                    cnt++
                    if (i == p) pPos = cnt
                    if (i == q) qPos = cnt
                }
            } else {
                create(n+1)
            }
        }
    }

    create(0)

    println(Math.abs(pPos - qPos))
}