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 <cmath>
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#include<set>
#include<tuple>
#include<bitset>
#include<iomanip>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
ll inf = 1000000007;
ll p[10];
ll q[10];
ll index[10];
int main(void) {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> p[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> q[i];
	}

	for (int i = 0; i < n; i++) {
		index[i] = i + 1;
	}
	int x, y;
		int count = 1;
	do {
		int check = 0;
		int check2 = 0;
		for (int i = 0; i < n; i++) {
			if (p[i] == index[i]) {
				check++;
			}
		}
		for (int i = 0; i < n; i++) {
			if (q[i] == index[i]) {
				check2++;
			}
		}
		if (check == n) {
			x = count;
		}
		if (check2 == n) {
			y = count;
		}

		count++;
	} while (next_permutation(index, index + n));
	cout << abs(x - y) << endl;
	return 0;
}