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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program count_order
  implicit none
  integer :: n, p(8) = 0, q(8) = 0, i
  integer :: a = 0, b = 0
  integer :: x(8) = (/1,2,3,4,5,6,7,8/)
  read(*,*) n
  read(*,*) p(1:n)
  read(*,*) q(1:n)
  call perm(x(1:n),[integer::])
  write(*,'(i0)') abs(a-b)
contains
  recursive subroutine perm(lin, lout)
    integer, intent(in) :: lin(:), lout(:)
    integer :: i
    if (size(lin) == 0) then
      a = a+cmp(p,lout)
      b = b+cmp(q,lout)
    else
      do i = 1, size(lin)
        call perm(pack(lin, lin /= lin(i)), [lout, lin(i)])
      end do
    end if
  end
  integer function cmp(x,y)
    integer, intent(in) :: x(:), y(:)
    integer :: i
    cmp = 0
    do i = 1, n
      if (x(i) > y(i)) then
        cmp = 1
        return
      else if (x(i) < y(i)) then
        return
      end if
    end do
  end
end program count_order