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.*;
public class Main {
	
    public static int fac(int n){
        if(n==0) return 1;
        return n*fac(n-1);
    }     
    public static void main(String[] args){        
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
		int [] in = new int [a];
        int [] in2 = new int [a];
        int b = 1;
        int c = 1;
        for(int i=0;i<a;i++){
                             in[i] = sc.nextInt();
                            }
        for(int i=0;i<a;i++){
                             in2[i] = sc.nextInt();
                            }
		for(int i=0;i<a;i++){b+=(in[i]-1)*fac(a-1-i);
        for(int j=0;j<a;j++){if(in[j]>in[i]){in[j]--;}                                                             
                            }                                                      
                            }  
        for(int i=0;i<a;i++){c+=(in2[i]-1)*fac(a-1-i);
        for(int j=0;j<a;j++){if(in2[j]>in2[i]){in2[j]--;}                                                             
                            }                                                      
                            }                                    
		System.out.println(Math.max(b-c,c-b));
	}
}