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;
#define rep(i, a, b) for (ll i = (a); i < (b); i++)
#define rep2(r, aa, bb) for (ll r = (aa)-1; r >= (bb); r--)
#define be(x) (x.begin(), x.end())
#define pb(t) push_back(t)
#define ll long long
#define mp(a, b) make_pair(a, b)
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n, count=0;
    cin>>n;
    vector<int>A(n), B(n);
    rep(i,0,n)cin>>A[i];
    rep(i,0,n)cin>>B[i];
    rep(i,0,n){
		if(A[i]<B[i]){
			break;	
		}else if(A[i]>B[i]){
					do{
				if(A!=B){
					count++;
				}else{
					cout<<count<<"\n";
					return 0;
				}
			}while(next_permutation(B.begin(), B.end()));
		}
	}
	do{
		if(A!=B){
			count++;
		}else{
			cout<<count<<"\n";
			return 0;
		}
	}while(next_permutation(A.begin(), A.end()));
}