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 Rust (1.15.1) (judge verdict: Accepted)

fn main () {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let p: Vec<_> = s.by_ref().take(n).map(|a| a.parse::<i32>().unwrap() - 1).collect();
	let q: Vec<_> = s.by_ref().take(n).map(|a| a.parse::<i32>().unwrap() - 1).collect();
	let fact = |i: u32| {
		if i == 0 { 1 }
		else {
			(2..i + 1).fold(1, |r, i| r * i)
		}
	};
	let f = |a: Vec<i32>| {
		let mut r: u32 = 0;
		let mut b: u32 = (1 << n) - 1;
		for a in a {
			b ^= 1 << a;
			r += (b & ((1 << a) - 1)).count_ones() * fact(b.count_ones());
		}
		r as i32
	};
	println!("{}", (f(p) - f(q)).abs());
}