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 Go (1.6) (judge verdict: Accepted)

package main

import "fmt"

func main() {
	p, q := input()
	a := sub(permitationNumber(p), permitationNumber(q))
	fmt.Println(a)
}

func sub(p, q int) int {
	if p < q {
		return q - p
	}
	return p - q
}

func permitationNumber(n []int) int {
	if len(n) <= 1 {
		return 0
	}
	return permitationNumber(n[1:]) + minNumber(n[0], n[1:])*factorial(len(n)-1)
}

func minNumber(m int, n []int) int {
	c := 0
	for _, v := range n {
		if v < m {
			c++
		}
	}
	return c
}

func factorial(n int) int {
	s := 1
	for ; n > 1; n-- {
		s *= n
	}
	return s
}

func input() ([]int, []int) {
	var n int
	fmt.Scan(&n)
	p := make([]int, n, n)
	q := make([]int, n, n)
	for i := range p {
		fmt.Scan(&p[i])
	}
	for i := range q {
		fmt.Scan(&q[i])
	}
	return p, q
}