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

fun main(args: Array<String>) {
    val N = readInputLine().toInt()
    val P = readInputLine().split(" ").map { it.toInt() }.toIntArray()
    val Q = readInputLine().split(" ").map { it.toInt() }.toIntArray()

    val perm = IntArray(N) { it + 1 }

    var cnt = 1
    var a = 0
    var b = 0

    do {
        var ok = true
        for (i in 0 until N) {
            if (P[i] != perm[i]) {
                ok = false
                break
            }
        }
        if (ok) a = cnt
        ok = true
        for (i in 0 until N) {
            if (Q[i] != perm[i]) {
                ok = false
                break
            }
        }
        if (ok) b = cnt
        cnt++
    } while (nextPermutation(perm))

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

fun readInputLine(): String {
    return readLine()!!
}

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;
}