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];
        int[] Q = new int[N];
        ArrayList<Integer> Pnumlist = new ArrayList<Integer>();
        ArrayList<Integer> Qnumlist = new ArrayList<Integer>();
        int[] fact = new int[N];

        fact[0] = 1;
        for(int i =1; i <N;i++){
            fact[i] = fact [i-1] * i;
        }
        for(int i=0;i<N;i++){
            Pnumlist.add(i+1);
            Qnumlist.add(i+1);
        }

        for(int i =0;i<N;i++){
            P[i] = sc.nextInt();
        }

        for(int i=0;i<N;i++){
            Q[i] = sc.nextInt();
        }

        int Pans =1;
        int Qans =1;
        for(int i=0;i<N;i++){
            Pans += fact[N-i-1] * (Pnumlist.indexOf(P[i]));
            Pnumlist.remove(Pnumlist.indexOf(P[i]));
            Qans += fact[N-i-1] * (Qnumlist.indexOf(Q[i]));
            Qnumlist.remove(Qnumlist.indexOf(Q[i]));

        }



        System.out.println(Math.abs(Pans-Qans));


    }

}