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>
#define rep(i, n) for(int i = 0; i < n; i++)
#define repr(i, n) for(int i = n; i >= 0; i--)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
int n;

int main()
{

  cin >> n;
  vector<int> p(n);
  vector<int> q(n);
  vector<int> seq(n);
  rep(i, n) cin >> p[i];
  rep(i, n) cin >> q[i];
  rep(i, n) seq[i] = i+1;
  map<vector<int>, int> mp;
  do {
    mp[seq] = mp.size();
  } while(next_permutation(seq.begin(),seq.end()));
  cout << abs(mp[p] - mp[q]) << endl;
  
  return 0;
}