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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

long long ara[12],fck[12],sck[12];

int main()
{

    long long n,m,a,b,c,d,e,i,j,p=1,fr=0,sc=0;  ara[0]=1;

    for(i=1;i<=9;i++)
    {
        p=p*i;  ara[i]=p;

    }

    scanf("%lld",&n);

    b=n;

    for(i=1;i<=n;i++)
    {
        scanf("%lld",&a);

        if(i<n)
        {
            e=0;

            for(j=1;j<=n;j++)
            {
                if(fck[j]==0)e++;

                if(a==j){ fck[j]=1; break; }
            }

            c=ara[b]/b;  b--;
            d=c*(e-1);
            if(fr==0)fr=d+1;
            else
                fr=fr+d;


        }


    }

    b=n;

     for(i=1;i<=n;i++)
    {
        scanf("%lld",&a);

        if(i<n)
        {
            e=0;

            for(j=1;j<=n;j++)
            {
                if(sck[j]==0)e++;

                if(a==j){  sck[j]=1; break; }
            }

             c=ara[b]/b;  b--;
            d=c*(e-1);
            if(sc==0)sc=d+1;
            else
                sc=sc+d;


        }


    }



    printf("%lld\n",abs(fr-sc));



    return 0;
}