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.List;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			List<Integer> p = new ArrayList<>(), q = new ArrayList<>();
			IntStream.range(0, n).forEach(i -> p.add(scanner.nextInt() - 1));
			IntStream.range(0, n).forEach(i -> q.add(scanner.nextInt() - 1));
			System.out.println(Math.abs(calcOrder(p) - calcOrder(q)));
		}
	}

	private static int calcOrder(List<Integer> p) {
		if (p.size() == 1) {
			return 0;
		}
		return p.get(0) * frac(p.size() - 1) + calcOrder(subList(p));
	}

	private static List<Integer> subList(List<Integer> list) {
		int first = list.get(0);
		list.remove(0);
		for (int i = 0; i < list.size(); i++) {
			Integer number = list.get(i);
			list.set(i, (number > first) ? number - 1 : number);
		}
		return list;
	}

	private static int frac(int n) {
		if (n <= 0) {
			return 1;
		}
		int result = 1;
		for (int i = 2; i <= n; i++) {
			result *= i;
		}
		return result;
	}
}