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)

fun main(args: Array<String>) {
    fun nextPermutation(arr: IntArray): Boolean {
        var i = arr.size - 1
        while (i > 0 && arr[i - 1] >= arr[i]) {
            i--
        }

        if (i <= 0) {
            return false
        }

        var j = arr.size - 1
        while (arr[j] <= arr[i - 1]) {
            j--
        }

        var temp = arr[i - 1]
        arr[i - 1] = arr[j]
        arr[j] = temp

        j = arr.size - 1
        while (i < j) {
            temp = arr[i]
            arr[i] = arr[j]
            arr[j] = temp
            i++
            j--
        }

        return true
    }

    val N = readLine()!!.toInt()

    val PN = readLine()!!.split(" ").map(String::toInt).toIntArray()
    val QN = readLine()!!.split(" ").map(String::toInt).toIntArray()

    val array = (1..N).toList().toIntArray()

    var index = 1
    var a = 0
    var b = 0
    do {
        if (PN.joinToString("") == array.joinToString("")) {
            a = index
        }
        if (QN.joinToString("") == array.joinToString("")) {
            b = index
        }
        index++

    } while (nextPermutation(array))

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