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.Scanner;

public class Main {
    private static int n, cnt = 0, x = 0;
    private static int[][] targetSeq = new int[2][10];
    private static int[] vis = new int[10];
    private static int[] tempSeq = new int[10];
    private static int[] ans = new int[2];

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        for(int i = 0;i < 2;i++) {
            for(int j = 0;j < n;j++) {
                targetSeq[i][j] = in.nextInt();
            }
        }
        dfs(0);
        System.out.println(Math.abs(ans[0] - ans[1]));
    }

    private static void dfs(int k) {
        if(x == 2) {
            return;
        }
        if(k == n) {
            cnt++;
            for(int i = 0;i < 2;i++) {
                boolean isOk = true;
                for(int j = 0;j < n;j++) {
                    if(tempSeq[j] != targetSeq[i][j]) {
                        isOk = false;
                        break;
                    }
                }
                if(isOk) {
                    ans[x++] = cnt;
                }
            }
            return;
        }
        for(int i = 1;i <= n;i++) {
            if(vis[i] == 0) {
                vis[i] = 1;
                tempSeq[k] = i;
                dfs(k + 1);
                vis[i] = 0;
            }
        }
    }
}