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.ArrayDeque;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Queue;
import java.util.Scanner;
 


public class Main {
	static int a1[]= {0,-1,0,1};
	static int b1[]= {-1,0,1,0};
	private static Scanner sc = new Scanner(System.in);
	//result = Math.pow(num1, num3)
	//StringBuffer str = new StringBuffer(hoge1);
	//String hoge2 = str.reverse().toString();
	//map.containsKey(A)

	//Map<String, Integer> map = new HashMap<String, Integer>(n);
	/*for ( キーのデータ型 key : マップの名前.keySet() ) {
		データのデータ型 data = マップの名前.get( key );
		
		// keyやdataを使った処理;
	}*/
	//int i = Integer.parseInt(s);
	//Queue<String> qq=new ArrayDeque<>(); //add,poll,peek
	//Deque<String> qq=new ArrayDeque<>();//push,pop,peek
	//ArrayList<String> cc = new ArrayList<>(n);
	//Collections.sort(list);
	//Array.sort(list);   cc.contains(tmp)
	//Arrays.asList(c).contains("a")
	//list.set(1,"walk");//1 1 2 3 5
	static int K,N;
	static int mod =1000000007;
	public static void main(String[] args) {
		int N=sc.nextInt();
		String a="";
		String b="";
		for(int i=0;i<N;i++) {
			a=a+sc.next();
		}
		for(int i=0;i<N;i++) {
			b=b+sc.next();
		}
		String f="";
		for(int i=1;i<=N;i++) {
			f=f+i;
		}
		String v=f;
		int ctn=1;
		int aV=0,bV=0;
		boolean aa=true,bb=true;
		if(a.equals(b)) {
			p(0);
		}else {
		while(aa||bb) {
			if(v.equals(a)) {
				aa=false;
				aV=ctn;
			}else if(v.equals(b)){
				bb=false;
				bV=ctn;
			}
			//p(v);
			v=nextPermutation(v);
			if(v.equals("Final")) {
				break;
			}
			ctn++;
		}
		p(Math.abs(aV-bV));
	}
		}

	
	//static ArrayList<Integer> cc = new ArrayList<>(10001);
	
	static void p(String ans) {System.out.println(ans);};
	static void p(int ans) {System.out.println(ans);};
	static void p() {System.out.println();};
	static void p(long ans) {System.out.println(ans);};
	static void p(double ans) {System.out.println(ans);};
	static String nextPermutation(String s) {
		ArrayList<Character> list=new ArrayList<>();
		for(int i=0;i<s.length();i++) {
			list.add(s.charAt(i));
		}
		int pivotPos=-1;
		char pivot=0;
		for(int i=list.size()-2;i>=0;i--) {
			if(list.get(i)<list.get(i+1)) {
				pivotPos=i;
				pivot=list.get(i);
				break;
			}
		}
		if(pivotPos==-1&&pivot==0) {
			return "Final";
		}
		int L=pivotPos+1,R=list.size()-1;
		
		
		
		int minPos=-1;
		char min =Character.MAX_VALUE;
		
		for(int i=R;i>=L;i--) {
			if(pivot<list.get(i)) {
				if(list.get(i)<min) {
					min=list.get(i);
					minPos=i;
				}
			}
		}
		
		Collections.swap(list, pivotPos, minPos);
		Collections.sort(list.subList(L, R+1));
		
		
		StringBuilder sb=new StringBuilder();
		for(int i=0;i<list.size();i++) {
			sb.append(list.get(i));
		}
		return sb.toString();
	}
}