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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static List<String> permutation(String list) {
        List<String> ans = new ArrayList<>();
        if (list.length() == 1) {
            return Collections.singletonList(list);
        }
        for (int i = 0; i < list.length(); i++) {
            char pick = list.charAt(i);
            String rest = list.substring(0, i) + list.substring(i + 1);
            for (String rest_permutation : permutation(rest)) {
                ans.add(pick + rest_permutation);
            }
        }
        return ans;
    }
    public static void main(String[] args) {
        //コード
        Scanner scanner = new Scanner(System.in);
        Integer N = Integer.parseInt(scanner.next());
        String[][] line = new String[2][1];
        for(int i = 0; i < 2; i++) {
            Arrays.fill(line[i], "");
            for(int j = 0; j < N; j++) {
                line[i][0] += scanner.next();
            }
        }
        String number = "";
        for(int i = 1; i <= N; i++) {
            number += i;
        }

        List<String> p = permutation(number);
        Integer R = 0;

        for (int j = 0; j < 2; j++) {
            for (int i = 0; i < p.size(); i++) {
                if(p.get(i).equals(line[j][0])) {
                    if(R==0)R+=i;
                    else R-=i;
                };
            }

        }
        System.out.println(Math.abs(R));
    }
}