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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.LinkedHashSet;

public class Main {
	static int[] p, q;
	static int cnt = 0;
	static int a, b;

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		sa = br.readLine().split(" ");
		p = new int[n];
		for (int i = 0; i < n; i++) {
			p[i] = Integer.parseInt(sa[i]);
		}
		sa = br.readLine().split(" ");
		q = new int[n];
		for (int i = 0; i < n; i++) {
			q[i] = Integer.parseInt(sa[i]);
		}
		br.close();

		int[] target = new int[n];
		for (int i = 0; i < n; i++) {
			target[i] = i + 1;
		}
		permutation(target, new LinkedHashSet<>());
		System.out.println(Math.abs(a - b));
	}

	static void permutation(int[] target, LinkedHashSet<Integer> set) {
		if (set.size() == target.length) {
			cnt++;
			boolean flg = true;
			int x = 0;
			for (int i : set) {
				if (target[i] != p[x]) {
					flg = false;
					break;
				}
				x++;
			}
			if (flg) {
				a = cnt;
			}

			flg = true;
			x = 0;
			for (int i : set) {
				if (target[i] != q[x]) {
					flg = false;
					break;
				}
				x++;
			}
			if (flg) {
				b = cnt;
			}
			return;
		}

		for (int i = 0; i < target.length; i++) {
			if (!set.contains(i)) {
				set.add(i);
				permutation(target, set);
				set.remove(i);
			}
		}
	}
}