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.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        Integer[] p = new Integer[n];
        Integer[] q = new Integer[n];
        for (int i = 0; i < n; i++) {
            p[i] = Integer.parseInt(sc.next());
        }
        for (int i = 0; i < n; i++) {
            q[i] = Integer.parseInt(sc.next());
        }
        Integer[] aa = new Integer[n];
        for (int i = 0; i < n; i++) {
            aa[i] = i + 1;
        }
        List<Integer[]> list = new ArrayList<>();
        Permutation perm = new Permutation();
        perm.rec(list, aa, new Integer[aa.length], 0);
        int a = 0;
        int b = 0;
        s: for (int i = 0; i < list.size(); i++) {
            for (int j = 0; j < n; j++) {
                if (list.get(i)[j] != p[j]) {
                    continue s;
                }
            }
            a = i;
            break;
        }
        s: for (int i = 0; i < list.size(); i++) {
            for (int j = 0; j < n; j++) {
                if (list.get(i)[j] != q[j]) {
                    continue s;
                }
            }
            b = i;
            break;
        }
        System.out.println(Math.abs(a - b));

    }

    private static class Permutation {

        public <T> void rec(List<T[]> list, T[] array, T[] ans, int index) {
            if (isEnd(array)) {
                list.add(ans.clone());
            } else {
                for (int i = 0; i < array.length; ++i) {
                    T o = array[i];
                    if (o == null) {
                        continue;
                    }
                    array[i] = null;
                    ans[index] = o;
                    rec(list, array, ans, index + 1);
                    ans[index] = null;
                    array[i] = o;
                }
            }
        }

        private <T> boolean isEnd(T[] array) {
            for (T o : array) {
                if (o != null) {
                    return false;
                }
            }
            return true;
        }
    }
}