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: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		// 入力値の取得
		int N = sc.nextInt();
		StringBuilder P = new StringBuilder();
		for (int i = 0; i < N; i++) {
			P.append(sc.nextInt());
		}
		StringBuilder Q = new StringBuilder();
		for (int i = 0; i < N; i++) {
			Q.append(sc.nextInt());
		}

		List<String> list = new ArrayList<>();
		permutation(list, "12345678".substring(0, N), "");
		
		int a = 0, b = 0;
		for (int i = 0; i < list.size(); i++) {
			if (list.get(i).equals(P.toString())) {
				a = i;
			}
			if (list.get(i).equals(Q.toString())) {
				b = i;
			}
		}

		// 結果の出力
		System.out.println(Math.abs(a - b));

		sc.close();
	}

	public static List<String> permutation(List<String> list, String target, String ans) {
		if (target.length() <= 1) {
			list.add(ans + target);
		} else {
			for (int i = 0; i < target.length(); i++) {
				permutation(list, target.substring(0, i) + target.substring(i + 1), ans + target.charAt(i));
			}
		}
		return list;
	}
}