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 cswork
{
    class Program
    {
        public Program() { }

        Scanner cin;
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<string> permList = new List<string>();
        void Omame()
        {
            cin = new Scanner();
            int n = cin.nextInt();

            string p = "";
            for (int i = 0; i < n; i++)
            {
                string pp = cin.next();
                p += pp + " ";
            }
            p = p.Substring(0, p.Length - 1);

            string q = "";
            for (int i = 0; i < n; i++)
            {
                string qq = cin.next();
                q += qq + " ";
            }
            q = q.Substring(0, q.Length - 1);

            Permutation(0, n, n);

            int a = 0;
            int b = 0;
            for (int i = 0; i < permList.Count; i++)
            {
                if (permList[i].Equals(p))
                {
                    a = i;
                }
                else if (permList[i].Equals(q))
                {
                    b = i;
                }
            }

            Console.WriteLine(Math.Abs(a - b));
        }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                string s = "";
                for (int i = 0; i < n; i++)
                {
                    s += (perm[i] + 1) + " ";
                }
                s = s.Substring(0, s.Length - 1);
                permList.Add(s);
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }
    }

    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }

        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();

            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next();
        }

        public int nextInt()
        {
            return int.Parse(next());
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }
    }
}