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: Wrong Answer)

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];

		int p = 0;
		int q = 0;
		int count = 0;

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

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

		for(int i = 0; i < n; i++) {
			if(!isMin(P, i)) {
				count += (P[i] - 1 - i) * factorial(P.length - (i+1));
			}
		}
		count += factorial(P.length - 2);

		p = count;
		count = 0;

		for(int i = 0; i < n; i++) {
			if(!isMin(Q, i)) {
				count += (Q[i] - 1 - i) * factorial(Q.length - (i+1));
			}
		}
		count += factorial(Q.length - 2);

		q = count;
		count = 0;

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

	public static int factorial(int n) {
		if(n > 1) {
			return factorial(n - 1) * n;
		}else {
			return n;
		}
	}

	public static boolean isMin(int[] list, int i) {
		for(int n = i + 1; n < list.length; n++) {
			if(list[i] > list[n]) {
				return false;
			}
		}
		return true;
	}
}