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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int kaizyou(int a)
{
    int ans = 1;
    for (int i = 0; i < a; i++)
    {
        ans *= a-i;
    }

    return ans;
    
}

int main()
{
   int N;
   cin >> N;
   vector<int> kazu(N);
   vector<int> kkazu(N);
   vector<int> P(N);
   vector<int> Q(N);
   for (int i = 0; i < N; i++)
   {
       cin >> P.at(i);
   }

   for (int i = 0; i < N; i++)
   {
       cin >> Q.at(i);
   }
   
   for (int i = 0; i < N; i++)
   {
       for (int j = i; j >= 0 ; j--)
       {
           if (P.at(i) > P.at(j))
           {
               kazu.at(i) ++;
           }

           if (Q.at(i) > Q.at(j))
           {
               kkazu.at(i) ++;
           }
           
           
       }
       
   }
   

   int a = 0;
   for (int i = 0; i < N; i++)
   {
       a += (P.at(i)-1-kazu.at(i))*kaizyou(N-1-i);
   }
   

   int b = 0;
   for (int i = 0; i < N; i++)
   {
       b += (Q.at(i)-1-kkazu.at(i))*kaizyou(N-1-i);
   }

   int ans;
   if (a >= b)
   {
       ans = a-b;
       cout << ans;
   }

   else
   {
       ans = b-a;
       cout << ans;
   }
   
   
   
   
}