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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
  public static int[] kai;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    kai = new int[9];
    kai[0] = 1;
    for(int i = 1; i < 9; i++) {
      kai[i] = (i * kai[i - 1]);
    }
    int n = sc.nextInt();
    ArrayList<Integer> listp = new ArrayList<Integer>();
    ArrayList<Integer> listq = new ArrayList<Integer>();
    for(int i = 0; i < n; i++) {
      listp.add(sc.nextInt());
    }
    for(int i = 0; i < n; i++) {
      listq.add(sc.nextInt());
    }
    int a = dfs(listp.get(0), listp);
    int b = dfs(listq.get(0), listq);
    System.out.println(Math.abs(a - b));
  }

  public static int dfs(int p, ArrayList<Integer> list) {
    int t = list.size();
    int ret = (p - 1) * kai[t - 1];
    if(t == 1) {
      return 1;
    } else {
      ArrayList<Integer> list2 = new ArrayList<Integer>();
      for(int i = 1; i < t; i++) {
        list2.add(list.get(i));
      }
      ArrayList<Integer> list3 = list2;
      Collections.sort(list3);
      int k = 0;
      for(int i = 0; i < (t - 1); i++) {
        if(list3.get(i) == list2.get(0)) {
          k = (i + 1);
          break;
        }
      }
      ret += dfs(k, list2);
      return ret;
    }
  }
}