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.Arrays;
import java.util.List;
import java.util.Scanner;

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

        for (int i = 0; i < n; i++) {
            q[i] = std.nextInt();
        }

        int[] sortedP = Arrays.copyOf(p, p.length);
        Arrays.sort(sortedP);
        int[] sortedQ = Arrays.copyOf(q, q.length);
        Arrays.sort(sortedQ);

        List<List<Integer>> listP = new ArrayList<>();
        backtrack(listP, new ArrayList<>(), sortedP);

        List<List<Integer>> listQ = new ArrayList<>();
        backtrack(listQ, new ArrayList<>(), sortedQ);

        int indexP = matchedIndex(listP, p);
        int indexQ = matchedIndex(listQ, q);
        int ans = Math.abs(indexP - indexQ);
        System.out.println(ans);
    }

    static int matchedIndex(List<List<Integer>> list, int[] nums) {
        int index = 0;
        for (int i = 0; i < list.size(); i++) {
            List<Integer> elements = list.get(i);
            boolean isMatched = true;
            for (int j = 0; j < elements.size(); j++) {
                int listValue = elements.get(j);
                if (listValue != nums[j]) {
                    isMatched = false;
                    break;
                }
            }

            if (isMatched) {
                index = i;
            }
        }

        return index;
    }

    static void backtrack(List<List<Integer>> out, List<Integer> tmpList, int[] nums) {
        if (tmpList.size() == nums.length) {
            out.add(new ArrayList<>(tmpList));
        } else {
            for (int i = 0; i < nums.length; i++) {
                if (tmpList.contains(nums[i])) {
                    continue;
                }

                tmpList.add(nums[i]);
                backtrack(out, tmpList, nums);
                tmpList.remove(tmpList.size() - 1);
            }
        }
    }
}