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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

int main(){
  int n;
  int *a,*b;
  int i;
  int fact=1;
  int a_num=0,b_num=0;
  int result;
  
  scanf("%d",&n);
  a=(int*)malloc(sizeof(int)*n);
  b=(int*)malloc(sizeof(int)*n);
  for(i=0;i<n;i++){
    scanf("%d",&a[i]);
  }
  for(i=0;i<n;i++){
    scanf("%d",&b[i]);
  }
  
  for(i=1;i<=n;i++){
    fact=fact*i;
  }
      
  for(i=n;i>0;i++){
    a_num+=(a[i]-1)*fact;
    b_num+=(b[i]-1)*fact;
    fact=fact/n;
  }
  
  result=abs(a_num-b_num);
  printf("%d",result);
  
  free(a);
  free(b);
}