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>
#define rep(i,n)for(int i=0;i<n;++i)
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int, int>P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//max=({});
//条件式が真ならwhileの中身を回し続ける
//printf("%d\n", ans);

//pairの入力
//vector<pair<ll, ll>>work(n);
//rep(i, n) {
//	ll a, b;
//	cin >> a >> b;
//	work[i] = make_pair(a, b);
//for(auto p:mp)(mapの探索)
//printf("%.10f\n",なんちゃら)

int g[15][15];
const int INF = 1001001001;
const int dx[4] = { -1,0,1,0 };
const int dy[4] = { 0,-1,0,1 };

//最大公約数
ll gcd(ll x, ll y) {
	return y ? gcd(y, x % y) : x;
}

ll lcm(ll x, ll y) {
	if (x == 0 || y == 0)return 0;
	return (x / gcd(x, y) * y);
}

//素因数分解
map<ll, ll>ins;
void facterize(ll n) {
	ll a = 2;
	while (n >= a * a) {
		if (n % a == 0) {
			ins[a]++;
			ins[a] = ins[a] % 1000000007;
			n /= a;
		}
		else {
			a++;
		}
	}
	ins[n]++;
}

vector<ll>relation[20], money(20, 1);
int dp[110][110];

int keta(int x) {
	int y = 0;
	while (x > 0) {
		y++;
		x /= 10;
	}
	return y;
}

int s[101][101];

int main() {
	int n;
	cin >> n;
	vector<int>p(n), q(n), r(n), r2(n);
	rep(i, n) {
		cin >> p[i];
		r[i] = i + 1;
	}
	rep(i, n) {
		cin >> q[i];
		r2[i] = i + 1;
	}
	int a = 0, b = 0, c = 1, d = 1;
	do {
		bool flag = true;
		rep(i, n) {
			if (p[i] != r[i])flag = false;
		}
		if (flag)a = c;
		c++;
	} while (next_permutation(r.begin(), r.end()));
	do {
		bool flag = true;
		rep(i, n) {
			if (q[i] != r2[i])flag = false;
		}
		if (flag)b = d;
		d++;
	} while (next_permutation(r2.begin(), r2.end()));
	cout << abs(a - b) << endl;
	return 0;
}