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: Wrong Answer)

using System;
using System.Collections.Generic;

namespace CSharp
{
    class Program
    {
        static void Main()
        {
            var N = int.Parse(Console.ReadLine());
            var P = Console.ReadLine().Replace(" ", "");
            var Q = Console.ReadLine().Replace(" ", "");

            var s = string.Empty;
            for (int i = 1; i < N + 1; i++)
            {
                s += i.ToString();
            }

            var l = Permutation(s);

            var pi = 0;
            var qi = 0;
            for (int i = 0; i < l.Count; i++)
            {
                if (l[i] == P)
                {
                    pi = i;
                }
                else if (l[i] == Q)
                {
                    qi = i;
                }
            }
            Console.WriteLine(Math.Abs(pi - qi));
        }

        static List<string> Permutation(string s)
        {
            if (s.Length == 1)
            {
                return new List<string>() { s };
            }
            var result = new List<string>();
            for (int i = 0; i < s.Length; i++)
            {
                foreach (var t in Permutation(s.Remove(i, 1)))
                {
                    result.Add(s[i] + t);
                }
            }
            return result;
        }
    }
}