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;
using LL = long long;
using PII = pair<int, int>;
using VI = vector<int>;

int main() {
#ifdef LOCAL_EXEC
//	freopen("sample.in", "r", stdin);
//	freopen("sample.out", "w", stdout);
#else
  ios_base::sync_with_stdio(false); cin.tie(NULL);
#endif

  int n; cin >> n;
  vector<vector<int>> perms;
  vector<int> perm(n);
  iota(perm.begin(), perm.end(), 1);
  do {
    perms.push_back(perm);
  } while (next_permutation(perm.begin(), perm.end()));

  vector<int> p(n), q(n);
  for (int& x: p) cin >> x;
  for (int& x: q) cin >> x;
  int ans = 0;
  for (int i = 0; i < (int) perms.size(); i++) {
    if (perms[i] == p) {
      ans -= i;
    }
    if (perms[i] == q) {
      ans += i;
    }
  }
  ans = abs(ans);
  cout << ans << endl;

  return 0;
}