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;
using ll = long long;
const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = ;
ll mod(ll A, ll M) {return (A % M + M) % M;}
const ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll divceil(ll A, ll B) {return (A + (B - 1)) / B;}
ll myctoi(char C) {return C - '0';}
char myitoc(ll N) {return '0' + N;}
#define FINALANS(A) {cout << (A) << '\n'; exit(0);}

int main()
{
  ll N;
  cin >> N;
  vector<ll> P(N), Q(N);
  for (ll i = 0; i < N; i++)
  {
    cin >> P.at(i);
  }
  for (ll i = 0; i < N; i++)
  {
    cin >> Q.at(i);
  }

  vector<ll> p(N);
  for (ll i = 0; i < N; i++)
  {
    p.at(i) = i + 1;
  }

  ll a = -1, b = -1;
  ll i = 1;
  do
  {
    bool ok = true;
    for (ll j = 0; j < N; j++)
    {
      if (p.at(j) != P.at(j))
      {
        ok = false;
        break;
      }
    }
    if (ok)
      a = i;

    ok = true;
    for (ll j = 0; j < N; j++)
    {
      if (p.at(j) != Q.at(j))
      {
        ok = false;
        break;
      }
    }
    if (ok)
      b = i;

    //cerr << a << " " << b << " " << i << endl;
    i++;
  } while (next_permutation(p.begin(), p.end()));

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