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.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] p = new int[n];
		int[] q = new int[n];

		for (int i = 0; i < n; i++) {
			p[i] = sc.nextInt();
		}
		for (int i = 0; i < n; i++) {
			q[i] = sc.nextInt();
		}

		System.out.println(Math.abs(searchCount(p) - searchCount(q)));
		sc.close();
	}

	public static int searchCount(int[] orderArr) {

		int len = orderArr.length;
		ArrayList<Integer> usedList = new ArrayList<Integer>();
		int count = 0;

		for (int i = 1; i <= len; i++) { // 桁数
			for (int j = 1; j <= len; j++) { // 注目している桁の値
				if ((orderArr[i - 1] > j) && (!usedList.contains(j))) {
					count += factorial(len - i);
				} else if (orderArr[i - 1] == j) {
					usedList.add(j); // 使った値を記録しておく
				}
			}
		}
		return count + 1;
	}

	public static int factorial(int n) {
		if (n == 0) {
			return 1;
		}
		return n * factorial(n - 1);
	}

}