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;
#define ll long long

int main(){
    vector<int>kai {0,1,2,6,24,120,720,5040,40320};
    int n,ansa = 0,ansb = 0;
    cin >> n;
    vector<int>va(n); vector<int>vb(n);
    for (int i = 0; i < n; i++) cin >> va.at(i);
    for (int i = 0; i < n; i++) cin >> vb.at(i);
    for (int i = 0; i < n-1; i++){
        int x = kai.at(n-1-i);
        ansa += x * (va.at(i)-1);
        ansb += x * (vb.at(i)-1);
        for (int j = 0; j < n; j++){
            if (va.at(i) < va.at(j)) va.at(j)--;
            if (vb.at(i) < vb.at(j)) vb.at(j)--;
        }
    }
    cout << abs(ansa - ansb) << "\n";
}