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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = (int)(a); i <= (int)(n); ++i)
#define rrep(i, a, n) for (int i = (int)(a); i >= (int)(n); --i)
#define debug(x) cerr << #x << " = " << x << "\n"
#define debugv(x) \
  rep(f, 0, (x.size() - 1)) cerr << x[f] << (f == (x.size() - 1) ? "\n" : " ")
#define debug2(x, y)                       \
  cerr << "(" << #x << "," << #y << ") = " \
       << "(" << x << "," << y << ")\n"
#define all(x) x.begin(), x.end()
#define each(a, x) for (auto& a : (x))
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define int long long
using namespace std;
using pii = pair<int, int>;
using vi = vector<int>;
const int MX = 1e5 + 5, INF = 5LL << 57, MOD = 1e9 + 7;

int N, M;
string S;
vi va, vb;

void input() {
  ;
  cin >> N;
  va.resize(N);
  vb.resize(N);
  rep(i, 0, N - 1) { cin >> va[i]; }
  rep(i, 0, N - 1) { cin >> vb[i]; };
}
void solve() {
  int cnt = 0;
  do {
    bool b = true;
    rep(i, 0, N - 1) {
      if (va[i] != vb[i]) {
        b = false;
      }
    }
    if (b) {
      break;
    }
    cnt++;

  } while (next_permutation(all(va)));
  int c1 = cnt;

  cnt = 0;

  do {
    bool b = true;
    rep(i, 0, N - 1) {
      if (va[i] != vb[i]) {
        b = false;
      }
    }
    if (b) {
      break;
    }
    cnt++;

  } while (next_permutation(all(vb)));
  int c2 = cnt;

  int fact = 1;
  int n = N;
  while (n > 1) {
    fact = fact * n;
    --n;
  }
  int a1 = abs(c1 - c2);

  ;
  cout << min(a1, fact - a1) << endl;
}
signed main() {
  input();
  solve();
  return 0;
}