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.lang.Math;
class Main{
  public static void main(String[] args){
    Scanner scan=new Scanner(System.in);
    int num=scan.nextInt();
    int a[]=new int[num];
    int b[]=new int[num];
    for(int x=0; x<num;x++){
      a[x]=scan.nextInt();
    }
    for(int x=0; x<num; x++){
      b[x]=scan.nextInt();
    }
    System.out.println(Math.abs(count(a,num)-count(b,num)));
  }
  public static int count(int array[], int num){
    int count=0, Ans=0;
    for(int x=0; x<array.length-1; x++){
      count=0;
      for(int y=x+1; y<array.length; y++){
        if(array[x]>array[y]){
          count++;
        }
      }
      System.out.println(count);
      Ans+=count*stair(num-x-1);
    }
    return Ans;
  }
  
  public static int stair(int in){
    int count=1;
    for(int x=in; x>1; x--){
      count*=x;
    }
    return count;
  }
}