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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
	static int count_p,count_q,p,q,n;
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		p = 0;
		q = 0;
		int mul = pow(10 , n - 1);
		for(int i = 0 ; i < n ; i++) {
			p += sc.nextInt() * mul;
			mul /= 10;
		}
		mul = pow(10 , n - 1);
		for(int i = 0 ; i < n ; i++) {
			q += sc.nextInt() * mul;
			mul /= 10;
		}
		count_p = 0;
		count_q = 0;
		count("");

		System.out.println(Math.abs(count_p - count_q));

	}
	public static void count(String s) {
		if(s.length() == n) {
			int i = Integer.valueOf(s);
			if(p >= i)
				count_p++;
			if(q >= i)
				count_q++;
			return;
		}
		for(int i = 1 ; i <= n ; i++) {
			String s2 = s;
			if(!s2.contains(i + "")) {
				count(s2 += i);
			}
		}

	}
	public static int pow(int a,int b) {
        int ret = 1;
        int tmp = a;
        while(b>0) {
            if((b&1)==1) ret = (ret * tmp);  //2進数によるべき乗の高速計算
            tmp *= tmp;
            b = b>>1;
        }
        return ret;
    }
}