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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e4+5;

int n,a,b,cnt;
int x[10],y[10],t[10];
 
inline bool check1(){
	for(int i = 0;i < n;++i){
		if(x[i] != t[i]) return 0;
	}
	a = cnt;
	return 1;
}

inline bool check2(){
	for(int i = 0;i < n;++i){
		if(y[i] != t[i]) return 0;
	}
	b = cnt;
	return 1;
}
 
int main()
{
    scanf("%d",&n);
    for(int i = 0;i < n;++i) scanf("%d",&x[i]),t[i] = i+1;
    for(int i = 0;i < n;++i) scanf("%d",&y[i]);
    
    cnt = 1;
    bool f1 = 0,f2 = 0;
    do{
    	if(!f1) check1();
    	if(!f2) check2();
    	if(f1 && f2) break;
    	cnt++;
	}while(next_permutation(t,t+n));
	
	printf("%d\n",abs(a-b));
    return 0;
}