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: Wrong Answer)

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

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
#define rep(i, s, n) for(int i = (int)(s); i < (int)(n); i++)

ll INF = 1ll << 60;

void re(int &cnt, vector<bool> used, vvi &per, int dps){
  int n = used.size()-1;
  rep(i, 0, n+1){
    if(used[i] == false)
      break;
    else if(i == n){
      cnt++;
      per[cnt] = per[cnt-1];
      return;
    }
  }
  
  rep(i, 1, n+1){
    if(used[i] == false){
      vector<bool> used_new = used;
      used_new[i] = true;
      per[cnt][dps] = i;
      re(cnt, used_new, per, dps+1);
    }
  }
}

int main(){
  int n;
  cin >> n;
  vi p(n), q(n);
  rep(i, 0, n)
    cin >> p[i];
  rep(i, 0, n)
    cin >> q[i];
  
  vvi per(50000, vi(n,0));
  vector<bool> used(n+1, false);
  used[0] = true;
  int cnt = 0;
  
  re(cnt, used, per, 0);
  
  int a, b;
  
  rep(i, 0, 50000){
    if(per[i] == p)
      a = i;
    if(per[i] == q)
      b = i;
  }
  
  cout << abs(a-b) << endl;
  
  return 0;
  
}