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<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<utility>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
using namespace std;
#define INF 2000001145
#define LINF 100000000000000007
#define MOD 1000000007
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define repb(i, n) for (int i = n - 1; i >= 0; i--)
#define MODE 1
#ifdef MODE
#define DEB(X) cout<< #X <<": "<<X<<" ";
#define ARDEB(i,X) cout<< #X <<"["<<i<<"]: "<<X[i]<<" ";
#define END cout<<endl;
#else
#define DEB(X) {}
#define ARDEB(i,X) {}
#define END {}
#endif
int ceil2(int a,int b){if(a%b){return a/b+1;}else{return a/b;}}
typedef long long ll;
typedef pair<int,int> P;
struct edge{int to,cost;};

int ans=0;
int k,n;
//string s;
int vk[111],sk[111];
vector<int> v,s;
signed main(){
	
	cin>>n;
	rep(i,n){
		int a;
		cin>>a;
		vk[i]=a;
		v.push_back(a);
	}
	rep(i,n){
		int a;
		cin>>a;
		sk[i]=a;
		s.push_back(a);
	}
	sort(v.begin(),v.end());
	sort(s.begin(),s.end());
	int ans2,cnt=0;
	do{
		cnt++;
		int f=1,f2=1;
		rep(i,n){
			if(vk[i]!=v[i])f=0;
			if(sk[i]!=s[i])f2=0;
		}
		if(f)ans=cnt;
		if(f2)ans2=cnt;
	}while(next_permutation(v.begin(),v.end())&&next_permutation(s.begin(),s.end()));
	cout<<abs(ans-ans2)<<endl;
}