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;
class Program
{
        private static int iN = 0;                                          // 順列の元の数
        private static int iK = 0;                                          // 順列の選ぶ数

    static void Main(string[] args)
    {
            // 入力
            string sS = System.Console.ReadLine().Trim();
            int iNN = int.Parse(sS);
            string[] sA = System.Console.ReadLine().Trim().Split(' ');
            string[] sB = System.Console.ReadLine().Trim().Split(' ');
            string ssA = "", ssB = "";
            for (int i = 0; i < sA.Length; i++) { ssA += (int.Parse(sA[i]) - 1).ToString(); }   // 入力文字列0始まりに変更スペース排除
            for (int i = 0; i < sB.Length; i++) { ssB += (int.Parse(sB[i]) - 1).ToString(); }   // 入力文字列0始まりに変更スペース排除

            System.Diagnostics.Stopwatch sw = new System.Diagnostics.Stopwatch();
            sw.Start();

            // 解析
            iN = iNN;
            iK = iNN;
            List<List<int>> lstCmp = new List<List<int>>();                 // 全順列(重複なし) 0始まり
            List<int> lstTmp = new List<int>();                             // テンポラリ
            GetCom(lstCmp, lstTmp);                                         // 全順列取得(重複なし)
            List<string> lstAll = new List<string>();                       // 全順列(重複なし) 0始まりの文字列
            for (int i = 0; i < lstCmp.Count; i++)
            {
                string sBuf = "";
                for (int j = 0; j < lstCmp[i].Count; j++)
                {
                    sBuf += lstCmp[i][j].ToString();
                }
                lstAll.Add(sBuf);                                           // 全順列(重複なし) 0始まりの文字列生成
            }
            int iAc = 0, iBc = 0;
            for (int i = 0; i < lstAll.Count; i++)
            {
                if (lstAll[i] == ssA) { iAc = i; }                          // 全順列内でのAのIndex位置 
                if (lstAll[i] == ssB) { iBc = i; }                          // 全順列内でのBのIndex位置
            }
            int iOut = Math.Abs(iAc - iBc);                                 // Index位置の差の絶対値

            // 出力
            System.Console.WriteLine("{0}", iOut.ToString().Trim());

            sw.Stop();
#if DEBUG
            Console.WriteLine();
            Console.WriteLine("{0}", sw.Elapsed.ToString());
            Console.WriteLine("続行するには何かキーを押してください．．．");
            Console.ReadKey();
#endif
    }

        // 重複なし順列
        private static void GetCom(List<List<int>> lstCmp, List<int> lstTmp)
        {
            for (int j = 0; j < iN; j++)
            {
                int iKn = 1;
                List<int> lstTm2 = new List<int>();
                lstTm2.AddRange(lstTmp);
                lstTm2.Add(j);
                GetCo2(lstCmp, lstTm2, iKn);
            }
        }

        // 順列2
        private static void GetCo2(List<List<int>> lstCmp, List<int> lstTmp, int iKn)
        {
            if (iKn == iK) { lstCmp.Add(lstTmp); return; }
            for (int j = 0; j < iN; j++)
            {
                if (lstTmp.Contains(j)) { continue; }
                List<int> lstTm2 = new List<int>();
                lstTm2.AddRange(lstTmp);
                lstTm2.Add(j);
                GetCo2(lstCmp, lstTm2, iKn + 1);
            }
        }

}