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)

/*                        ______
 _______   /\     |``\   |        |  /
    |     /  \    |__/   |____    |/
    |    / _ _\   |  \   |        |\
    |   /      \  |   \  |______  |  \

            Dept. of CSE
        Comilla  University

*/
#include<bits/stdc++.h>
#define pi 2*acos(0.0)
#define ll long long
#define FIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define Node struct node
#define spc " "
#define  E  2.71828182845904523536
#define pb push_back
#define pp pop_back
#define valid(nx,ny) nx>=0 && nx<n && ny>=0 && ny<m
#define ee endl
#define pii pair<int,int>
#define infinity 10000000000000000
#define mod 1000000007
int fx[]={0,0,+1,-1};
int fy[]={+1,-1,0,0};

using namespace std;
///check n=1;
///check corner case;
const int maX=1e5+1;
vector<ll>Graph[maX],cost[maX];
ll visit[maX];


int main()
{
   ll n,i,j,cc=0,k;
   cin>>n;
   ll p[n],q[n];for(i=0;i<n;i++)cin>>p[i];
   for(i=0;i<n;i++)cin>>q[i];
   vector<ll>a;
   for(i=0;i<n;i++)a.pb(i+1);
   ll pp=0,qq=0,x=0,y=0;
   while(1){
     ll cc=0;
      for(i=0;i<n;i++){
        if(p[i]!=a[i]){cc=1;break;}
      }
      if(cc==1 && x==0)pp++;
      else x=1;
      ll dd=0;
      for(i=0;i<n;i++){
        if(q[i]!=a[i]){dd=1;break;}
      }
      if(dd==1 && y==0)qq++;
      else y=1;
      if(x && y)break;
      next_permutation(a.begin(),a.end());
   }
   cout<<abs(pp-qq)<<endl;
return 0;
}