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>

int fact[] = {1,1,2,6,24,120,720,5040,40320};
int N;
int flag[]= {1,1,1,1,1,1,1,1,1};
int getcount(int x)
{
	int count=0;
	flag[x] = 0;
	for(int i=1;i<x;i++)
		count+=flag[i];
	return count;
}
int getrank(int P[])
{
	int j,rank=1;
	j=1;
	while(j<N)
	{
		rank += getcount(P[j])*fact[N-j];
		j++;
	} 
	return rank;
}
int abs(int x)
{
	if (x<0)
		return x*-1;
	else
		return x;
}

int main()
{
	scanf("%d",&N);
	int P[N+1];
	int Q[N+1];
	int i,rankp,rankq;
	for(i=1;i<=N;i++)
		scanf("%d",&P[i]);
	for(i=1;i<=N;i++)
		scanf("%d",&Q[i]);
	rankp = getrank(P);
	for(i=1;i<=N;i++)
		flag[i] = 1;
	rankq = getrank(Q);
	printf("%d",abs(rankp-rankq));


	
}