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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
typedef long long int ll;

int main()
{
    int n;
    cin >> n;
    vector<int>p(n),q(n);
    for(int i = 0; i < n; i++) cin >> p[i];
    for(int i = 0; i < n; i++) cin >> q[i];
    if(p == q) 
    {
        cout << 0 << endl;
        return 0;
    }
    int k = 0;
    int r;
    while(1)
    {
        if(p[k] == q[k]) k++;
        if(p[k] > q[k]) 
        {
            r = 0; //pの方が辞書順では大きい
            break;
        }
        if(p[k] < q[k]) 
        {
            r = 1; //qの方が辞書順では大きい
            break;
        }
    }
    int ans = 0;
    if(r == 0)
    {
        do
        {
            bool f = true;
            for(int i = 0; i < n; i++) if(p[i] != q[i]) f = false;
            if(!f) ans++;
            else break;
        }while(next_permutation(q.begin(),q.end()));
    }
    if(r == 1)
    {
        do
        {
            bool f = true;
            for(int i = 0; i < n; i++) if(p[i] != q[i]) f = false;
            if(!f) ans++;
            else break;
        }while(next_permutation(p.begin(),p.end()));
    }
    cout << ans << endl;
}