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, n) for(int i = 0; i < (int)(n); i++)
using namespace std;

int factorial(int num) {
  int sum = 1;
  for (int i = 1; i <= num; i++) sum *= i;
  return sum;
}

int conutOrder(vector<int> permutation) {
  int cnt = 1;
  rep (i, permutation.size() - 1) {
    int checkNum = i + 1;
    if (permutation[i] != checkNum) {
      cnt += factorial(permutation.size() - 1) * (permutation[i] - 1);
    } else {
      if (permutation[i] < checkNum) {
        cnt += i - 1;
      } else {
        cnt += i;
      }
    }
  }
  return cnt;
}

int main() {
  int n;
  cin >> n;
  vector<int> p(n);
  rep (i, n) {
    cin >> p[i];
  }
  vector<int> q(n);
  rep (i, n) {
    cin >> q[i];
  }

  int a, b;
  a = conutOrder(p);
  b = conutOrder(q);

  cout << abs(a - b) << endl;
  return 0;
}