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;

public class Main {
	static int p;
	static int q;
	static int n;
	static int a = -1;
	static int b = -1;
	static int cnt = 0;
	
	static void dfs(String v) {
		if (v.length() == n) {
//			System.out.print(v);
			cnt++;
			if (Integer.valueOf(v) == p) {
				a = cnt;
//				System.out.printf(" a cnt=%d", cnt);
			}
			if (Integer.valueOf(v) == q) {
				b = cnt;
//				System.out.printf(" b cnt=%d", cnt);
			}
//			System.out.println();
			return;
		}
		
		for (int i = 1; i <= n; i++) {
			if (v.contains(Integer.toString(i))) {
				continue;
			}
			String nv = v + i;
			dfs(nv);
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		
		p = sc.nextInt();
		for (int i = 1; i < n; i++) {
			p = p * 10 + sc.nextInt();
		}
		
		q = sc.nextInt();
		for (int i = 1; i < n; i++) {
			q = q * 10 + sc.nextInt();
		}
		
//		System.out.printf("p=%d\n", p);
//		System.out.printf("q=%d\n", q);
		
		for (int i = 1; i <= n; i++) {
			dfs(Integer.toString(i));
		}
		
		int ans = Math.abs(a - b);
		System.out.println(ans);
	}
}