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: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		exec2();
	}

	private static void exec() {
		try (Scanner sc = new Scanner(System.in)) {
			int N = sc.nextInt();
			
			long M = sc.nextLong();

			long l = 1;
			
			for (int i = 0; i < N; i++) {
				long in = (long)sc.nextLong()/2;
				l = lcm(l, in);
			}
			
			if(l>M) {
				System.out.println(0);
				return;
			}
			
			long ans = 0;
			if((M/l)%2==0) {
				ans = M/l/2;
			} else {
				ans = M/l/2 + 1;
			}
			
			System.out.println(ans);
		}
	}
	
	private static void exec2() {
		try (Scanner sc = new Scanner(System.in)) {
			int N = sc.nextInt();
			
			int[] a = new int[N];
			int[] b = new int[N];
			
			for (int i = 0; i < N; i++) {
				a[i] = sc.nextInt();
				
			}
			for (int i = 0; i < N; i++) {
				b[i] = sc.nextInt();
			}
			
			int anum =1;
			for (int i = 0; i < N; i++) {
					for(int j=i+1;j<N;j++) {
						if(a[i] < a[j]) {
							a[j]--;
						}
				}
				anum += p((N-1-i))*(a[i]-1);
			}
			
			int bnum =1;
			for (int i = 0; i < N; i++) {
					for(int j=i+1;j<N;j++) {
						if(b[i] < b[j]) {
							b[j]--;
						}
				}
				bnum += p((N-1-i))*(b[i]-1);
			}
			
			System.out.println(anum);
			System.out.println(bnum);
			System.out.println(Math.abs(anum-bnum));
		}
	}
	
	private static int p(int a) {
		if(a==1) return 1;
		if(a<1) return 0;
		return a*p(a-1);
	}
	
	private static long gcd(long m, long n) {
        if(m < n) return gcd(n, m);
        if(n == 0) return m;
        return gcd(n, m % n);
    }
    
    private static long lcm(long m, long n) {
        return m * n / gcd(m, n);
    }
}