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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define mod 1000000007
#define dbg(n) cout<<#n<<' '<<n<<endl;
#define dbg_v(v) cout<<#v<<":";for(int i=0;i<(int)v.size();i++) cout<<" "<<v[i]; cout<<endl;
int main(){ 
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);	
	int t=1;
	//cin>>t;
	while(t--){
		int n;
		cin>>n;
		int p[n];
		int q[n];
		int a[n];
		for (int i = 0; i < n; i++)
		{
			cin>>p[i];
			a[i]=i+1;
		}
		for (int i = 0; i < n; i++)
		{
			cin>>q[i];
		}
		int i=1,c,b;
		do{
			int f=0,g=0;
			for(int i=0;i<n;i++){
				if(a[i]!=p[i])
					f=1;
			}
			for(int i=0;i<n;i++){
				if(a[i]!=q[i])
					g=1;
			}
			if(!f)
				c=i;
			if(!g)
				b=i;
			i++;
		}while (next_permutation(a,a+n));
		cout<<abs(c-b)<<endl; 
	}
}