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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <climits>
#include <cmath>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define INF_INT (INT_MAX / 2)
#define INF_LONG (LONG_MAX / 2)
//#define DEBUG true
#define DEBUG false
using namespace std;

const int MAX = 100001;
const int MOD = 1000000007;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

int ceil(int x, int y) { return (x % y == 0) ? x / y : x / y + 1; }

int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }

int lcm(int x, int y) { return x / gcd(x, y) * y; }

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  int p[n], q[n];
  for (int i = 0; i < n; i++) cin >> p[i];
  for (int i = 0; i < n; i++) cin >> q[i];

  int perm[n];
  for (int i = 0; i < n; i++) perm[i] = i + 1;

  int a = 0;
  do {
    bool ok = true;
    for (int i = 0; i < n; i++) {
      if (perm[i] != p[i]) ok = false;
    }
    if (ok) break;
    ++a;
  } while (next_permutation(perm, perm + n));

  for (int i = 0; i < n; i++) perm[i] = i + 1;
  int b = 0;
  do {
    bool ok = true;
    for (int i = 0; i < n; i++) {
      if (perm[i] != q[i]) ok = false;
    }
    if (ok) break;
    ++b;
  } while (next_permutation(perm, perm + n));

  int ans = abs(a - b);

  cout << ans << endl;

  return 0;
}