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;
import java.util.Collections;
import java.util.List;
import java.util.ArrayList;
public class Main {
    public static void main (String[] args) {
    Scanner sc = new Scanner(System.in);
	int N = sc.nextInt();
    List<Integer> P = new ArrayList<Integer>();
    List<Integer> Q = new ArrayList<Integer>();
    List<Integer> Ps = new ArrayList<Integer>();
    List<Integer> Qs = new ArrayList<Integer>();   
    int a = 0; 
    int b = 0;
    for(int i=0; i<N; i++){
    	int x = sc.nextInt();
        P.add(x);
        Ps.add(x);
    }
    for(int i=0; i<N; i++){
    	int x = sc.nextInt();
        Q.add(x);
        Qs.add(x);
    } 
    Collections.sort(Ps);
    Collections.sort(Qs);
      
   for(int i=0; i<N-1;i++){
     int searchNumber = P.get(i);
     int x = 0;
     int y = 1;
     for(int j=0; j<Ps.size(); j++){
     	if(searchNumber == Ps.get(j)){
          x = j;
          Ps.remove(j);
          break;
        }
     }
   for( int k = (N - i - 1); k > 0; k--) y *= k;
   if(x != 0) a += x * y;
   }
      
   for(int i=0; i<N-1;i++){
     int searchNumber = Q.get(i);
     int x = 0;
     int y = 1;
     for(int j=0; j<Qs.size(); j++){
     	if(searchNumber == Qs.get(j)){
          x = j;
          Qs.remove(j);
          break;
        }
     }
   for( int k = (N - i - 1); k > 0; k--) y *= k;
   if(x != 0) a += x * y;
   }
      
	if(a == 0) a =1;
    if(b == 0) b =1;
      
   int answer = a - b;
   if(answer < 0) answer = -answer;
   System.out.println(answer);   
    }
}