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>

long fac(long n)
{
	long ans=1;
	int i;
	
	if(n==0 || n==1)
		return 1;
	
	for(i=1;i<=n;i++)
		ans *= i;
	return ans;
}

long num(long n, int a[], int i)   //returns order
{
	int j;
	int count=0;
	long ans;
	
	if(i==n)
		return 0;
	
	for(j=0;j<n-i;j++)
		if(a[i]>a[i+j])
			count++;
	ans = count*fac(n-i-1) + num(n,a,i+1);
	return ans;
}
	
	


int main(void)
{
	int n;
	int i;
	int a[8]={0};
	int b[8]={0};
	
	scanf("%d",&n);
	
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	
	for(i=0;i<n;i++)
		scanf("%d",&b[i]);
	
	printf("%ld",labs(num(n,a,0)-num(n,b,0)));
	return 0;
}