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

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];
		IntStream.rangeClosed(1, n).forEach(i -> p[i - 1] = sc.nextInt());
		IntStream.rangeClosed(1, n).forEach(i -> q[i - 1] = sc.nextInt());
		int[] convertedP = convert(p);
		int[] convertedQ = convert(q);
		int diff = 0;
		for (int i = 0; i < n; i++) {
			diff = diff + (convertedP[i] - convertedQ[i]) * pow(n - i - 1);
		}
		print(abs(diff));
		sc.close();
	}

	public static int[] convert(int[] n) {
		int[] converted = new int[n.length];
		for (int i = 0; i < n.length; i++) {
			converted[i] = n[i];
			int order = n[i];
			for (int j = 0; j < i; j++) {
				if (n[j] < order) {
					converted[i] = converted[i] - 1;
				}
			}
		}
		return converted;
	}

	public static int abs(int n) {
		return n > 0 ? n : -n;
	}

	public static int pow(int n) {
		int powNum = 1;
		for (int i = 1; i <= n; i++) {
			powNum = powNum * i;
		}
		return powNum;
	}

	public static void print(Object obj) {
		System.out.println(obj);
	}

}