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;

namespace Practice
{
    class Program
    {
        static void Main(string[] args)
        {
            //ここから
            int n = int.Parse(Console.ReadLine());
            string[] str = new string[n];
            int[] p = new int[n];
            int[] q = new int[n];

            str = Console.ReadLine().Split(' ');
            for(int i = 0; i < n; i++)
            {
                p[i] = int.Parse(str[i]);
            }

            str = Console.ReadLine().Split(' ');
            for (int i = 0; i < n; i++)
            {
                q[i] = int.Parse(str[i]);
            }

            int def = Math.Abs(jisho(n, p) - jisho(n, q));
            Console.WriteLine(def);
            


            //ここまで
        }

        static int jisho(int n, int[] x)
        {
            if(n == 1) { return 1; }
            else
            {
                int[] temp = new int[n - 1];
                for(int i = 0; i < n-1; i++)
                {
                    if (x[i + 1] > x[0]) { temp[i] = x[i + 1] - 1; }
                    else { temp[i] = x[i + 1]; }
                }
                return (x[0] - 1) * kaijo(n - 1) + jisho(n - 1, temp);
            }
        }
        static int kaijo(int n)
        {
            if (n == 1) { return 1; }
            else { return n * kaijo(n - 1); }
        }
    }
}