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.*;

public class Main {

    static <T extends Comparable<? super T>> boolean nextPermutation(T[] sequence) {
        int first = getFirst(sequence);
        if (first == -1) return false;
        int toSwap = sequence.length - 1;
        while (sequence[first].compareTo(sequence[toSwap]) >= 0) --toSwap;
        swap(sequence, first++, toSwap);
        toSwap = sequence.length - 1;
        while (first < toSwap) swap(sequence, first++, toSwap--);
        return true;
    }

    static <T extends Comparable<? super T>> int getFirst(T[] sequence) {
        for (int i = sequence.length - 2; i >= 0; --i)
            if (sequence[i].compareTo(sequence[i + 1]) < 0) return i;
        return -1;
    }

    static <T extends Comparable<? super T>> void swap(T[] sequence, int i, int j) {
        T tmp = sequence[i];
        sequence[i] = sequence[j];
        sequence[j] = tmp;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        Integer[] p = new Integer[n];
        Integer[] q = new Integer[n];
        Integer[] perm = new Integer[n];
        for (int i = 0; i < n; ++i) p[i] = sc.nextInt();
        for (int i = 0; i < n; ++i) q[i] = sc.nextInt();
        for (int i = 0; i < n; ++i) perm[i] = i + 1;

        int pNum = 0, qNum = 0;
        int cnt = 1;
        do {
            if (Arrays.equals(perm, p)) pNum = cnt;
            if (Arrays.equals(perm, q)) qNum = cnt;
            ++cnt;
        } while (nextPermutation(perm));

        System.out.println(Math.abs(pNum - qNum));
    }
}