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;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
void Yes() {cout << "Yes" << endl;}
void No() {cout << "No" << endl;}
void YES() {cout << "YES" << endl;}
void NO() {cout << "NO" << endl;}
void neg() {cout << - 1 << endl;}
void unhap() {cout << ":(" << endl;}

int main() {
  int n;
  cin >> n;
  vector<int> p(n), q(n);
  for (int i = 0; i < n; i++) cin >> p.at(i);
  for (int i = 0; i < n; i++) cin >> q.at(i);
  ll aa = 1, bb = 1;

  for (int j = 0; j < n - 1; j++) {
    ll a = 1, b = 1;
    for (int i = 1; i <= n - 1 - j; i++) {
      a *= i;
      b *= i;
    }
    aa += a * (p.at(j) - 1);
    bb += b * (q.at(j) - 1);
    for (int i = j + 1; i < n; i++) {
      if (p.at(i) > p.at(j)) p.at(i)--;
      if (q.at(i) > q.at(j)) q.at(i)--;
    }
  }
  cout << max(aa - bb, bb - aa) << endl;
}