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"
	"math"
	"sort"
)

func main() {
	var N int
	fmt.Scan(&N)
	a := make([]int, N)
	b := make([]int, N)
	for i := 0; i < N; i++ {
		fmt.Scan(&a[i])
	}
	for i := 0; i < N; i++ {
		fmt.Scan(&b[i])
	}
	x := make([]int, N)
	for i := 0; i < N; i++ {
		x[i] = i + 1
	}
	var ac, bc int
	arr := NextPermutation(x)
	for i := 0; i < len(arr); i++ {
		aeq := true
		beq := true
		for j := 0; j < N; j++ {
			if arr[i][j] != a[j] {
				aeq = false
				break
			}
		}
		if aeq {
			ac = i + 1
		}
		for j := 0; j < N; j++ {
			if arr[i][j] != b[j] {
				beq = false
				break
			}
		}
		if beq {
			bc = i + 1
		}
	}
	fmt.Println(Abs(ac - bc))
}
func Abs(a int) int {
	return int(math.Abs(float64(a)))
}

func NextPermutation(x []int) [][]int {
	var p [][]int
	z := make([]int, len(x))
	copy(z, x)
	p = append(p, z)
	for i := 1; NextPermutationSub(sort.IntSlice(x)); i++ {
		y := make([]int, len(x))
		copy(y, x)
		p = append(p, y)
	}
	return p
}

func NextPermutationSub(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}