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 void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt();
        int[] p=new int[n],q=new int[n];
        int[] facts={1,1,2,6,24,120,720,5040,40320};
        boolean[] usedp=new boolean[n];
        boolean[] usedq=new boolean[n];
        int pin=0;
        int qin=0;
        for(int i=0;i<n;i++)p[i]=sc.nextInt();
        for(int i=0;i<n;i++)q[i]=sc.nextInt();
        int countp;
        int countq;
        for(int i=0;i<n;i++){
            countp=0;
            countq=0;
            for(int j=0;j<n;j++){
                if(!usedp[j]&&j!=p[i]-1){
                    countp++;
                }else if(j==p[i]-1){
                    break;
                }
            }
            for(int j=0;j<n;j++){
                if(!usedq[j]&&j!=q[i]-1){
                    countq++;
                }else if(j==q[i]-1){
                    break;
                }
            }
            usedp[p[i]-1]=true;
            usedq[q[i]-1]=true;
            pin+=facts[n-i-1]*countp;
            
            qin+=facts[n-i-1]*countq;
        }
        System.out.println(Math.abs(pin-qin));
 
    }


}