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.List;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();

        final List<Integer> tmp = IntStream.rangeClosed(1, n)
                .boxed()
                .collect(Collectors.toList());

        int index = 1;
        for (int i = 0; i < n; i++) {
            final int p = scanner.nextInt();
            final int pIndex = tmp.indexOf(p);
            if (pIndex != i) {
                index += kaizyou(n - i) * (pIndex - i);
            }
            tmp.remove(pIndex);
        }

        final List<Integer> hoge = IntStream.rangeClosed(1, n)
                .boxed()
                .collect(Collectors.toList());

        int fuga = 1;
        for (int i = 0; i < n; i++) {
            final int p = scanner.nextInt();
            final int pIndex = hoge.indexOf(p);
            if (pIndex != i) {
                fuga += kaizyou(n - i) * (pIndex - i);
            }
            hoge.remove(pIndex);
        }

        System.out.println(Math.abs(fuga - index));
    }

    private static int kaizyou(final int i) {
        int result = 1;
        for (int j = 2; j < i; j++) {
            result *= j;
        }

        return result;
    }
}