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 <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;

int calc(int i)
{
	if (i <= 1)
		return 1;
	else
		return i * calc(i - 1);
}

int main()
{
	long long n, ans = 0,a=0,b=0;
	cin >> n;
	vector<long long> p(n),q(n);
	vector<bool> checkP(n, false), checkQ(n,false);
	for (long long i = 0; i < p.size(); i++)
		cin >> p[i];

	for (long long i = 0; i < q.size(); i++)
		cin >> q[i];

	for (long long i = 0; i < n-1; i++)
	{
		long long countP = 0,countQ=0;
		for (long long j = 0; j < p[i]; j++)
		{
			if (checkP[j])
				countP++;
		}
		for (long long j = 0; j < q[i]; j++)
		{
			if (checkQ[j])
				countQ++;
		}
		checkP[p[i]-1] = true;
		checkQ[q[i]-1] = true;
		a += (p[i] - countP) * calc(n - 1 - i);
		b += (q[i] - countQ) * calc(n - 1 - i);
	}

	ans = abs(a - b);

	cout << ans << endl;
}