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 Pascal (FPC 3.0.4) (judge verdict: Accepted)

// credit: fwho1150
program main;

type
    sample = array[1..8] of longint;

var
    n, order: longint;
    p, q: sample;

procedure input (var a: sample);
var
    i: longint;

begin
    for i := 1 to n do
        read(a[i]);
    readln;
end;

function code(var a: sample): sample;
var 
    i, j, c: longint;

begin
    for i := 1 to n do
        a[i] := a[i] - 1;
    for i := 1 to n do begin
        c := 0;
        for j := 1 to (i - 1) do
            if (a[j] < a[i]) then 
                inc(c);
        code[i] := a[i] - c;
    end;
end;

function fact(n: longint): longint;
begin
    if (n = 0) or (n = 1) then 
        fact := 1
    else 
        fact := n * fact(n - 1);
end;

function rank(a: sample): longint;
var
    i, tmp: longint;

begin
    tmp := 0;
    for i := 1 to n do 
        tmp := tmp + a[i] * fact(n - i);
    rank := tmp;
end;

begin
    readln(n);
    input(p);
    input(q);
    order := abs(rank(code(p)) - rank(code(q)));
    writeln(order);
end.