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: Wrong Answer)

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

int main()
{
    ll num;
    cin>>num;
    vector<ll> p(num), q(num), temp(num);
    for (int i = 0; i < num; ++i) {
        cin>>p[i];
        temp[i] = p[i];
    }
    for (int i = 0; i < num; ++i) {
        cin>>q[i];
    }
    sort(temp.begin(), temp.end());
    ll counter = 1;
    vector<ll> result;
    if(p == temp || q == temp)
    {
        if(p == q)
        {
            result.push_back(counter);
            result.push_back(counter);
        }
        else
            result.push_back(counter);
    }
    while (next_permutation(temp.begin(), temp.end()))
    {
        if(p == temp || q == temp)
        {
            if(p == q)
            {
                result.push_back(counter);
                result.push_back(counter);
            }
            else
                result.push_back(counter);
        }
        if(result.size() == 2)
            break;
        counter++;
    }
    cout<<result[1] - result[0];
    return 0;
}