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<math.h>
int perm(int n,int arr[])
{
    int arr1[n];
    int sum=0,count,i,j;
    int arr2[9]={1,1,2,6,24,120,720,5040,40320};
    for(i=0;i<n;i++)
    {
        for(j=i+1;j<n;j++)
        {
            if(arr[j]<arr[i])
            count++;
        }
        arr1[i]=count;
	count=0;
    }
    for(i=0;i<n;i++)
    {
        sum=sum+(arr1[i]*arr2[n-i-1]);
    }
    return sum+1;
}
int main()
{
    int n,res1,res2;
    scanf("%d",&n);
    int arr[n];
    int arr1[n];
    int i,j;
    for(i=0;i<=n-1;i++)
    {
        scanf("%d",&arr[i]);
    }
    for(i=0;i<=n-1;i++)
    {
        scanf("%d",&arr1[i]);
    }
    res1=perm(n,arr);
    res2=perm(n,arr1);
    printf("\n%d",abs(res1-res2));
    return 0;
}