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 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define for(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
int read(){
	int res=0,f=1;
	char c=getchar();
	if(c=='-') f=-1;
	while(c>'9'||c<'0') c=getchar();
	while(c>='0'&&c<='9'){
		res=res*10+c-'0';
		c=getchar();
	}
	return res*f;
}
int n;
vector<int> v;
vector<string> ans;
int main(){
	n=read();
	for(i,1,n) v.push_back(i);
	while(next_permutation(v.begin(),v.end())){
		string s="";
		s+=(char)(v[0]+48);
		for(i,1,v.size()-1) s=s+' '+(char)(v[i]+48);
		ans.push_back(s);
	}
	string s1,s2;
	getline(cin,s1);
	getline(cin,s2);
	if(s1==s2){
		cout<<0<<"\n";
		return 0;
	}
	int a,b;
	for(i,0,ans.size()-1){
		if(s1==ans[i]) a=i;
		if(s2==ans[i]) b=i;
	}
	cout<<abs(a-b)<<"\n";
	return 0;
}