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.io.BufferedWriter;
import java.io.FileWriter;
import java.io.InputStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigDecimal;
import java.math.MathContext;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

import static java.util.Comparator.*;

public class Main {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver(System.in, out);
        solver.solve();
        out.close();
    }
}
class Solver {
	Scanner sc;
	PrintWriter out;
    public Solver(InputStream in, PrintWriter out) {
    	sc = new Scanner(in);
    	this.out = out;
    }
    // ======================================================================
    public void solve() {
    	int N = ni();
    	int[] P = ndi(N);
    	int[] Q = ndi(N);
    	int[] no = new int[N];
    	for (int i = 0; i < N; i++) {
    		no[i] = i + 1;
		}
    	int a = -1, b = -1, cnt = 1;
    	boolean f;
    	do {
    		if(a == -1) {
    			f = true;
				for (int i = 0; i < N; i++) {
					if(P[i] != no[i]) {
						f = false;
					}
				}
				if(f)	a = cnt;
    		}
    		if(b == -1) {
    			f = true;
				for (int i = 0; i < N; i++) {
					if(Q[i] != no[i]) {
						f = false;
					}
				}
				if(f)	b = cnt;
    		}
    		cnt++;
    	} while (next_permutation(no));
    	out.println(Math.abs(a - b));
    }
	// 与えられた配列の中身を辞書順に、次の順列をセットする（次がなければ　false を返す）								
	public boolean next_permutation(int[] a) {								
		for (int i = a.length - 2; i >= 0; i--) {							
			if (a[i] < a[i + 1]) {						
				for (int j = a.length - 1;; j--) {					
					if (a[i] < a[j]) {				
						int temp = a[i];			
						a[i] = a[j];			
						a[j] = temp;			
						for (i++, j = a.length - 1; i < j; i++, j--) {			
							temp = a[i];		
							a[i] = a[j];		
							a[j] = temp;		
						}			
						return true;			
					}				
				}					
			}						
		}							
		return false;							
	}								
    // ------------------------------------------
    // 入力
    // ------------------------------------------
    public int ni() {
    	return sc.nextInt();
    }
    public long nl() {
    	return sc.nextLong();
    }
    public String ns() {
    	return sc.next();
    }
    public char[] nc() {
    	return sc.next().toCharArray();
    }
    public int[] ndi(int N) {
    	int[] ans = new int[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ni();
		}
    	return ans;
    }
    public long[] ndl(int N) {
    	long[] ans = new long[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = nl();
		}
    	return ans;
    }
    public String[] nds(int N) {
    	String[] ans = new String[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ns();
		}
    	return ans;
    }
    public char[][] ndc(int N) {
    	char[][] ans = new char[N][];
    	for (int i = 0; i < N; i++) {
			ans[i] = nc();
		}
    	return ans;
    }
    public int[][] nddi(int N, int S) {
    	int[][] ans = new int[N][S];
    	for (int i = 0; i < N; i++) {
			for (int j = 0; j < S; j++) {
				ans[i][j] = ni();
			}
		}
    	return ans;
    }
}