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)

#define rep(i, n) for(int i=0; i<n; i++)
#include <bits/stdc++.h>
using namespace std;

int fac(int k) {
	int sum = 1;
	for (int i = 1; i <= k; i++) {
		sum *= i;
	}
	return sum;
}

int main(void) {
	int n; cin >> n;
	vector<int> P(n), Q(n);
	rep(i, n) cin >> P[i];
	rep(i, n) cin >> Q[i];

	int a = 0, b = 0;
	for (int i = 0; i < n-1; i++) {
		if (P[i] != 1) {
			a += (P[i] - 1) * fac(n - i - 1);
		}
		if (Q[i] != 1) {
			b += (Q[i] - 1) * fac(n - i - 1);
		}
		for (int j = i + 1; j < n; j++) {
			if (P[i] < P[j]) {
				P[j]--;
			}
			if (Q[i] < Q[j]) {
				Q[j]--;
			}
		}
	}

	cout << abs(a - b) << endl;
}