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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <stdio.h>
#include <cmath>
#include <cctype>
#include <map>

const int INT_INF = 1000000000;                 // 10^9
const long long LONG_INF = 1000000000000000000; // 10^18

using namespace std;

int main()
{
    int n;
    cin >> n;

    vector<int> p(n), q(n), a(n);
    for (int i = 0; i < n; i++)
        cin >> p[i];
    for (int i = 0; i < n; i++)
        cin >> q[i];
    for (int i = 0; i < n; i++)
        a[i] = i + 1;

    int count = 1;
    map<vector<int>, int> m;
    while (true)
    {
        m[a] = count;
        count++;
        if (!next_permutation(a.begin(), a.end()))
        {
            break;
        }
    }

    int ans = abs(m[p] - m[q]);
    cout << ans << endl;
    return 0;
}