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;
class Program
{
  static void Main(string[] args)
  {
    // 整数の入力
    int N = int.Parse(Console.ReadLine());
  
    // スペース区切りの整数の入力
    int[] P = new int[N];
    
    string[] inputP = Console.ReadLine().Split(' ');
    for(int i = 0; i<N; i++){
      P[i] = int.Parse(inputP[i]);
    }

    int[] Q = new int[N];
    
    string[] inputQ = Console.ReadLine().Split(' ');
    for(int i = 0; i<N; i++){
      Q[i] = int.Parse(inputQ[i]);
    }

    //

    int temp = 0;

    int indexP = 0;
    
    for(int i = 0; i < N; i++){
      for(int j = 0; j < N; j++){
        if(P[i] == j+1){
          break;
        }else{
          for(int k=1; k<=(N-1-i); k++){
            if(k==1)
            {
              temp = 1;
            } else {
            temp *= k;
            }
          }
          indexP += temp;
        }
      }
    }

    int indexQ = 0;

    temp = 0;
    
    for(int i = 0; i < N; i++){
      for(int j = 0; j < N; j++){
        if(Q[i] == j+1){
          break;
        }else{
          for(int k=1; k<=(N-1-i); k++){
            if(k==1)
            {
              temp = 1;
            } else {
            temp *= k;
            }
          }
          indexQ += temp;
        }
      }
    }
    
  
    //出力
    Console.WriteLine(System.Math.Abs(indexP - indexQ));
  }
}