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;
typedef long long ll;
#define ff first
#define ss second
#define pb push_back
#define I insert
int main()
{
  
  ll N;
  cin>>N;
  ll i=1;
  vector<ll>A;
  vector<ll>B;
  vector<ll>C;
  for(i=0;i<N;i++)
    A.push_back(i+1);
  ll in;
  for(i=0;i<N;i++)
  {
    cin>>in;
    B.push_back(in);
  }
  for(i=0;i<N;i++)
  {
    cin>>in;
    C.push_back(in);
  }
  if(B==C)
    cout<<0;
  else
  {
    ll cnt1=0,cnt2=0;
    while(A!=B)
    {
      cnt1++;
     next_permutation(A.begin(),A.end());
    }
    A.clear();
    for(i=0;i<N;i++)
    {
    A.push_back(i+1);
    }
    while(C!=A)
    {
      cnt2++;
      next_permutation(A.begin(),A.end());
    }
    cout<<abs(cnt2-cnt1);
  }
    
  
  
}