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;

class CountOrder
{
    static void Main(string[] args)
    {

        var input =int.Parse( Console.ReadLine());

        var x = Enumerable.Range(1, input).ToArray();
        var p = AllPermutation(x);

        string[] start = Console.ReadLine().Split(' ');
        string[] end = Console.ReadLine().Split(' ');

        var a = 0;
        for(int i = 0;i < p.Count; i++)
        {
            var ok = true;
            for(int j = 0; j < input; j++)
            {
                if (int.Parse(start[j]) != p[i][j]) ok = false;
            }
            if (ok)
            {
                a = i;
                break;
            }
        }

        var b = 0;
        for (int i = 0; i < p.Count; i++)
        {
            var ok = true;
            for (int j = 0; j < input; j++)
            {
                if (int.Parse(end[j]) != p[i][j]) ok = false;
            }
            if (ok)
            {
                b = i;
                break;
            }
        }


        Console.WriteLine("{0}",Math.Abs(a-b));
    }

    static List<T[]> AllPermutation<T>(params T[] array) where T : IComparable
    {
        var a = new List<T>(array).ToArray();
        var res = new List<T[]>();
        res.Add(new List<T>(a).ToArray());
        var n = a.Length;
        var next = true;
        while (next)
        {
            next = false;

            // 1
            int i;
            for (i = n - 2; i >= 0; i--)
            {
                if (a[i].CompareTo(a[i + 1]) < 0) break;
            }
            // 2
            if (i < 0) break;

            // 3
            var j = n;
            do
            {
                j--;
            } while (a[i].CompareTo(a[j]) > 0);

            if (a[i].CompareTo(a[j]) < 0)
            {
                // 4
                var tmp = a[i];
                a[i] = a[j];
                a[j] = tmp;
                Array.Reverse(a, i + 1, n - i - 1);
                res.Add(new List<T>(a).ToArray());
                next = true;
            }
        }
        return res;
    }

}