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, to) for (int i = 0; i < (to); i++)
#define rrep(i, to) for (int i = to - 1; i >= 0; i--)
#define repf(i, from, to) for (int i = (from); i < (to); i++)
#define all(v) v.begin(), v.end()
#define unless(cond) if (!(cond))
using namespace std;
using ll = long long;
template <typename T>
using V = vector<T>;
template <typename T, typename U>
using P = pair<T, U>;

int main() {
  int n;
  V<int> p;
  V<int> q;

  cin >> n;
  p.resize(n);
  q.resize(n);
  rep(i, n) {
    cin >> p[i];
  }
  rep(i, n) {
    cin >> q[i];
  }

  V<int> perm(n, 0);
  rep(i, n) {
    perm[i] = i + 1;
  }

  bool find_p = false;
  bool find_q = false;
  int a, b;
  int count = 0;
  do {
    count++;
    if(perm == p && !find_p) {
      a = count;
    }
    if(perm == q && !find_q) {
      b = count;
    }
  } while(next_permutation(all(perm)));

  cout << abs(a - b) << endl;
}