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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Text;//for StringBuilder
class c
{
    static void Main()
    {
        int n = int.Parse(Console.ReadLine());
        var p = string.Join("", Array.ConvertAll(Console.ReadLine().Split(), int.Parse));
        var q = string.Join("", Array.ConvertAll(Console.ReadLine().Split(), int.Parse));
        var src = new List<int>();
        for (int i = 1; i <= n; i++) src.Add(i);
        var x = Permutations(src, n);
        var list = new List<string>();
        foreach (var item in x) list.Add(string.Join("", item));
        Console.WriteLine(Math.Abs(list.IndexOf(p) - list.IndexOf(q)));
    }
    static List<List<T>> Permutations<T>(List<T> src, int n)
    {
        // nは1以上かつsrcの長さよりも短い必要がある。
        int len = src.Count;
        if (!(0 < n && n <= len))
        {
            throw new ArgumentOutOfRangeException();
        }

        var permutations = new List<List<T>>();
        var queue = new Queue<List<int>>(Enumerable.Range(0, len).Select(i => new List<int>() { i }));
        while (queue.Any())
        {
            var indexes = queue.Dequeue();

            if (indexes.Count == n)
            {
                permutations.Add(indexes.Select(index => src[index]).ToList());
                continue;
            }

            for (int index = 0; index < len; index++)
            {
                if (indexes.Contains(index))
                {
                    continue;
                }

                var newIndexes = new List<int>(indexes);
                newIndexes.Add(index);
                queue.Enqueue(newIndexes);
            }
        }

        return permutations;
    }
}