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 Python (3.4.3) (judge verdict: Accepted)

def kaijou(n):
  result = 1
  for i in range(1, n+1):
    result *= i
  return result

def nanbanme(l, n):
  for i in range(len(l)):
    if l[i] == n:
      return i
  
n = int(input())
p = list(map(int, input().split()))
q = list(map(int, input().split()))
p_n = [i for i in range(1, n+1)]
q_n = [i for i in range(1, n+1)]

pp, qq = 1, 1

for i in range(n):
  if p[i] != min(p_n):
    pp += kaijou(n-i-1) * nanbanme(p_n, p[i])
  p_n.remove(p[i])
    
  if q[i] != min(q_n):
    qq += kaijou(n-i-1) * nanbanme(q_n, q[i])
  q_n.remove(q[i])
    
print(abs(pp-qq))