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;

int temp(int N) {
  if (N==0) return 1;
  return N*temp(N-1);
}

// index取得
int get_index(vector<int> vec, int x) {
  int index;
  for (int i = 0; i < distance(vec.begin(), vec.end()); i++) {
    //cout << vec.size() << endl;
    if (x == vec.at(i)) {
      index = i;
      return index;
    }
  }
  return -1;
}


int main() {
  // 入力
  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);
  }
  
  // 昇順の配列
  vector<int> arr_a(N), arr_b(N);
  for (int i = 0; i < N; i++) {
    arr_a.at(i) = i+1;
    arr_b.at(i) = i+1;
  }
  
  long a = 0;
  long b = 0;
  int x = 0;
  for (int i = 0; i < N-1; i++) {
    x = get_index(arr_a, P.at(i));
    arr_a.erase(arr_a.begin() + x);
    a += temp(N-1-i) * x;
  }
  
  for (int i = 0; i < N-1; i++) {
    x = get_index(arr_b, Q.at(i));
    arr_b.erase(arr_b.begin() + x);
    b += temp(N-1-i) * x;
  }
  
  if (a-b >= 0) cout << a-b << endl;
  else cout << b-a << endl;
  
  return 0; 
}