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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {

    private static List<Integer> permList = new ArrayList<>();

    private static void permutation(int n, Set<Integer> set, int num) {
        if (String.valueOf(num).length() >= n) {
            permList.add(num);
            return;
        }
        for (int i = 1; i <= n; i++) {
            if (set.contains(i)) {
                continue;
            }
            set.add(i);
            permutation(n, set, num * 10 + i);
            set.remove(i);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] p = new int[n];
        int[] q = new int[n];
        for (int i = 0; i < n; i++) {
            p[i] = sc.nextInt();
        }
        for (int i = 0; i < n; i++) {
            q[i] = sc.nextInt();
        }

        int pnum = 0;
        int qnum = 0;
        for (int i = 0; i < n; i++) {
            pnum += p[i] * Math.pow(10, n - i - 1);
            qnum += q[i] * Math.pow(10, n - i - 1);
        }

        permutation(n, new HashSet<>(), 0);
        int a = 0;
        int b = 0;
        for (int i = 0; i < permList.size(); i++) {
            if (pnum == permList.get(i)) {
                a = i;
            }
            if (qnum == permList.get(i)) {
                b = i;
            }
        }
        System.out.println(Math.abs(a - b));
    }

}