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 OCaml (4.02.3) (judge verdict: Accepted)

open Batteries
let rec perm n ls =
  if n = 0 then [[]]
  else flatmap (fun x -> List.map (fun y -> x :: y)
                                  (perm (n - 1) (remove x ls)))
               ls
and remove x ls = List.filter (fun y -> x <> y) ls
and flatten = function
  [] -> []
| x::xs -> x @ (flatten xs)
and flatmap func ls = flatten (List.map func ls)


let n = read_int ()
let rec m lst =
  match lst with
   [] -> ""
  | first :: rest -> first ^ m rest
let p = m (read_line () |> String.split_on_char ' ')
let q = m (read_line () |> String.split_on_char ' ')
let rec g k =
  match k with
   0 -> []
  | _ -> k :: g (k-1)
let all = 
  let rec h lst =
    match lst with
     [] -> []
    | first :: rest -> m(List.map string_of_int first) :: h rest
  in List.sort compare (h(perm n (g n)))

let _ = 
  let rec o target l s =
    match l with
     [] -> 0
    | first :: rest -> if first = target then s else o target rest (s+1)
  in let jj t = if t < 0 then t * (-1) else t
    in print_endline(string_of_int (jj( (o p all 1) - (o q all 1))))