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: Accepted)

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

int order(int P[8], int index, int N)
{
  	int	i;
  	int	count;	
  
  	count = 1;
  	for (i=index+1; i<N; i++)
    {
      	if (P[index] > P[i])
          	count++;
    }
  	return (count);
}

int	main(void)
{
  	int	N;
  	int P[8];
  	int Q[8];
  	int R[] = {5040,720,120,24,6,2,1,0};
 	int num;
  	int i;
  	int PP;
  	int QQ;
  
  	num = scanf("%d", &N);
    if (num != 1)
    {
      	return (0);
    }
  
  	for (i=0; i<N; i++)
    {
      	num = scanf("%d", &P[i]);
     	if (num != 1)
          	return (0);
    }
  	for (i=0; i<N; i++)
    {
      	num = scanf("%d", &Q[i]);
      	if (num != 1)
          	return (0);
    }
  	for (i=N; i<8; i++)
    {
      	P[i] = 0;
      	Q[i] = 0;
    }
  	
  	PP = 0;
  	for (i=0; i<N; i++)
    {
      	PP += order(P, i, N) * R[i + 8 - N];
    }
  	QQ = 0;
  	for (i=0; i<N; i++)
    {
      	QQ += order(Q, i, N) * R[i + 8 - N];
    }
  	printf("%d", abs(QQ - PP));
  	return (0);
}