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.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		ArrayList<Integer> listA = new ArrayList<>();
		ArrayList<Integer> listB = new ArrayList<>();
		
		for(int i=0;i<N;i++) {
			listA.add(sc.nextInt());
		}
		for(int i=0;i<N;i++) {
			listB.add(sc.nextInt());
		}
		int countA=1;
		int countB=1;
		for(int i=0;i<N;i++) {
			countA+=(countorder(listA.get(0),listA)-1)*factor(N-i-1);
			listA.remove(0);
		}
		for(int i=0;i<N;i++) {
			countB+=(countorder(listB.get(0),listB)-1)*factor(N-i-1);
			listB.remove(0);
		}
		System.out.println(Math.abs(countA-countB));
	}
	
	
	static int factor(int n) {
		if(n==0) {
			return 1;
		}else {
			return n*factor(n-1);
		}
	}
	
	static int countorder(int a,ArrayList<Integer> list) {
		int count=1;
		for(int i=0;i<list.size();i++) {
			if(list.get(i)<a) {
				count++;
			}
		}
		return count;
	}
}