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(4):: n, i
    integer(4), allocatable:: p(:), q(:)
    integer(4), parameter:: bi(8) = [1, 2, 6, 24, 120, 720, 5040, 40320]

    read*, n
    allocate(p(n), q(n))
    read*, p(:)
    read*, q(:)

    do i = 1, n-1
        where(p(i+1:) > p(i))
            p(i+1:) = p(i+1:) - 1
        end where
        where( q(i+1:) > q(i))
            q(i+1:) = q(i+1:) - 1
        end where
        p(i) = p(i) * bi(n-i)
        q(i) = q(i) * bi(n-i)
    end do

    print*, abs(sum(p-q))
end program count_order