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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

public class PaizaPj
{
    

    public static void Main()
    {
        // 読込
        int N = int.Parse(Console.ReadLine().Trim());
        var P = Console.ReadLine().Trim().Split(' ').ToList();
        var Q = Console.ReadLine().Trim().Split(' ').ToList();

        var origPerm = new List<string>(P);
        origPerm.Sort();

        string p = "";
        string q = "";
        string orig = "";
        for (int i = 0; i < N; i++)
        {
            p += P[i];
            q += Q[i];
            orig += origPerm[i];
        }

        var perms = Perms(orig);

        int pIdx = perms.FindIndex(s => s.Equals(p));
        int qIdx = perms.FindIndex(s => s.Equals(q));

        Console.WriteLine(Math.Abs(pIdx - qIdx));

        //Console.ReadLine();
    }

    static List<string> Perms(string s)
    {
        if (s.Length == 1) return new List<string>() { s };
        // ↑は以下でもよい
        // if (s.Length == 0) return new List<string>() { "" };

        var ret = new List<string>();
        for (int i = 0; i < s.Length; i++)
        {
            // 1. s[i] を除いたグループで順列を作る(再帰)
            foreach (string t in Perms(s.Remove(i, 1)))
            {
                // 2. 先頭に s[i] をつけると
                //    s[i] が先頭の順列が出来る(長さ s.Length の順列)
                ret.Add(s[i] + t);
            }
        }
        return ret;
    }


}