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: Wrong Answer)

//include
//------------------------------------------
#include <bits/stdc++.h>
using namespace std;

//typedef
//------------------------------------------
typedef long long LL;
typedef vector<LL> VL;
typedef vector<VL> VVL;
typedef vector<string> VS;
typedef pair<LL, LL> PLL;

//container util
//------------------------------------------
#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)
#define rwhole(f,x,...) ([&](decltype((x)) whole) { return (f)(rbegin(whole), rend(whole), ## __VA_ARGS__); })(x)
#define EACH(i,c) for(decltype((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) whole(sort, c)
#define RSORT(c) rwhole(sort, c)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int MOD = 1000000007;

// grid
//--------------------------------------------
VL dx = {0, 1, 0, -1};
VL dy = {1, 0, -1, 0};
VL dx2 = {-1, 0, 1, -1, 1, -1, 0, 1};
VL dy2 = {-1, -1, -1, 0, 0, 1, 1, 1};

//debug
//--------------------------------------------
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

//IO accelerate
//--------------------------------------------
struct InitIO {
    InitIO() {
        cin.tie(nullptr);
        ios_base::sync_with_stdio(false);
        cout << fixed << setprecision(30);
    }
} init_io;

//template
//--------------------------------------------
template<typename T>
istream& operator >>(istream& is, vector<T>& vec) {
  for(T& x: vec) is >> x;
  return is;
}
template<typename T>
ostream& operator <<(ostream& os, const vector<T>& vec) {
  for(int i=0; i<vec.size(); i++){
    os << vec[i] << ( i+1 == vec.size() ? "" : "\t" );
  }
  return os;
}
template<typename T>
ostream& operator <<(ostream& s, const vector<vector<T>>& vv) {
	for (int i = 0; i < vv.size(); ++i) {
		s << vv[i] << endl;
	}
	return s;
}

// 多重vector
// auto dp=make_v<int>(4,h,w) みたいに使える
template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

// 多重vectorのためのfill
// fill_v(dp,0) みたいに使える
template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
  for(auto &e:t) fill_v(e,v);
}

template<class T,class U> void chmax(T&a,U b){if(a<b)a=b;}
template<class T,class U> void chmin(T&a,U b){if(b<a)a=b;}

//main code

int main(int argc, char *argv[])
{
	int n;
	cin >> n;
	VL p(n), q(n);
	cin >> p >> q;
	VL a(n);
	for (int i = 0; i < n; i++) {
		a[i] = i+1;
	}
	int x,y;
	LL c = 0;
	do {
		if (a == p) {
			x = c;
		} else if (a == q) {
			y = c;
		}
		c++;
	} while(whole(next_permutation, a));
	cout << abs(x-y) << endl;
	return 0;
}