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 Ruby (2.3.3) (judge verdict: Wrong Answer)

def factorial (num)
  return 0 if num < 1
  (1..num).inject(&:*)
end

N = gets.strip.to_i
P = gets.strip.split(' ').map(&:to_i)
Q = gets.strip.split(' ').map(&:to_i)
p = 0
q = 0
overp = 0
overq = 0
N.times do |n|
  overp += 1 if n > 0 && P[n] > P[n-1]
  overq += 1 if n > 0 && Q[n] > Q[n-1]
  p1 = (P[n] - overp - n) * factorial(N - n - 1)
  q1 = (Q[n] - overq - n) * factorial(N - n - 1)
  p += p1
  q += q1
end
puts (p - q).abs