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<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<bitset>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define brep(index,num) for(int index=num-1;index>=0;index--)
#define brep1(index,num) for(int index=num;index>0;index--)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-7
#define mp(a1,a2) make_pair(a1,a2)
#define ALL(a) (a).begin(),(a).end()
#define rALL(a) (a).rbegin(),(a).rend()
typedef long long ll;
typedef long double ld;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int main(){
	int N,P[9],Q[9];
	scan(N);
	int pa=1,qa=1;
	vint A(N);
	rep(i,N){
		A[i]=i;
	}
	rep(i,N){
		scan(P[i]);
		P[i]--;
	}
	rep(i,N){
		scan(Q[i]);
		Q[i]--;
	}
	while(1){
		int flag=1;
		rep(i,N){
			if(A[i]!=P[i]){
				flag=0;
			}
		}
		if(flag){
			break;
		}
		pa++;
		next_permutation(ALL(A));
	}
	rep(i,N){
		A[i]=i;
	}
	while(1){
		int flag=1;
		rep(i,N){
			if(A[i]!=Q[i]){
				flag=0;
			}
		}
		if(flag){
			break;
		}
		qa++;
		next_permutation(ALL(A));
	}
	prin(abs(pa-qa));
	return 0;
}