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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define sim template <class c
#define ris return *this
#define dor > debug &operator<<
#define eni(x)                                                                        \
	sim > typename enable_if<sizeof dud<c>(0) x 1, debug &>::type operator<<(c i) \
	{
sim > struct rge
{
	c b, e;
};
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c *x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug
{
#ifdef LOCAL
	~debug()
	{
		cerr << endl;
	}
	eni(!=) cerr << boolalpha << i;
	ris;
} eni(==) ris << range(begin(i), end(i));
}
sim, class b dor(pair<b, c> d)
{
	ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d)
{
	*this << "[";
	for (auto it = d.b; it != d.e; ++it)
		*this << ", " + 2 * (it == d.b) << *it;
	ris << "]";
}
#else
	sim dor(const c &)
	{
		ris;
	}
#endif
}
;
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define forn(i, n) for (ll i = 0; i < int(n); i++)
#define rfor(n, i) for (ll i = (n - 1); i >= 0; --i)
#define all(V) V.begin(), V.end()
#define rall(V) V.rbegin(), V.rend()
#define len(V) (int)V.size()
#define ll long long
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define endl '\n'
#define M_PI 3.14159265358979323846
#define EPS 1e-7

void solve()
{
	int n; cin >> n;
	vector<int> a(n), ac(n), b(n), bc(n);
	forn(i, n) 
	{
		cin >> a[i];
		ac[i] = a[i];
	}
	forn(i, n) 
	{
		cin >> b[i];
		bc[i] = b[i];
	}
	sort(all(a));
	sort(all(b));
	ll cnt = 0;
	bool ok = false;
	do {
		ok = true;
		for (int i = 0; i < n; ++i)
		{
			if (a[i] == ac[i]) continue;
			else 
			{
				ok = false;
			}
		}
		cnt++;
	} while (next_permutation(all(a)) && !ok);
	ll cnt2 = 0;
	ok = true;
	do
	{
		ok = true;
		for (int i = 0; i < n; ++i)
		{
			if (b[i] == bc[i])
				continue;
			else
			{
				ok = false;
				break;
			}
		}
		cnt2++;
	} while (next_permutation(all(b)) && !ok);
	cout << abs(cnt - cnt2) << endl;
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	solve();
}