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++14 (GCC 5.4.1) (judge verdict: Accepted)

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

int main()
{
    int n,i,j,k;cin>>n;
    int p[n],q[n];
    for(i=0;i<n;i++) cin>>p[i];
    for(i=0;i<n;i++) cin>>q[i];
    int arr[n]; for(i=0;i<n;i++) arr[i]=i+1;
    k=1;int a,b;
    do
    {
        int f=1;
        for(i=0;i<n;i++)
        {
            if(arr[i]!=p[i])
            {
                f=0;break;
            }
        }
        if(f==1) a=k;

        f=1;
        for(i=0;i<n;i++)
        {
            if(arr[i]!=q[i])
               {
                   f=0;break;
               }
        }
        if(f==1) b=k;
        k++;
    }while(next_permutation(arr,arr+n));

    cout<<abs(a-b);




    return 0;
}