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)

// https://atcoder.jp/contests/abc150/tasks/abc150_c
// AtCoder Beginner Contest 150
// C - Count Order

/*
Problem.
大きさNの順列について、並びかえたP,Qが与えられる。
Pが辞書順a番目、Qが辞書順b番目として|a-b|は？

Point.
next_permutationを使えば辞書順に順列を生成してくれる。
*/

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<ctime>

using namespace std;
using ll = long long;

bool match(int n, vector<int> v, vector<int> p){
  for(int i = 0; i < n; i++){
    if(v[i] != p[i]) return false;
  }
  return true;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start-----------------------------------------------
  int n; cin >> n;
  vector<int> v(n);
  for(int i = 0; i < n; i++) v[i] = i+1;
  vector<int> p(n);
  vector<int> q(n);
  for(auto &i:p) cin >> i;
  for(auto &i:q) cin >> i;
  int a = 0, b = 0, cnt = 0;
  do{
    cnt++;
    if(match(n, v, p)) a = cnt;
    if(match(n, v, q)) b = cnt;
  }while(next_permutation(v.begin(), v.end()));
  cout << abs(a - b) << endl;
  // end-----------------------------------------------
  cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}