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 read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn rep<T, F: FnMut() -> T>(n: usize, mut f: F) -> Vec<T> {
    (0..n).map(|_| f()).collect()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let p: Vec<i32> = rep(n, || scan(tokens));
    let q: Vec<i32> = rep(n, || scan(tokens));
    let ord_p = order(p);
    let ord_q = order(q);
    println!("{}", (ord_p - ord_q).abs());
}

fn order(p: Vec<i32>) -> i32 {
    let mut x = p.clone();
    x.sort();
    let mut count = 0;
    for i in p {
        for j in 0..x.len() {
            if x[j] == i {
                x.remove(j);
                if j > 0 {
                    count += j * factorial(x.len());
                }
                break;
            }
        }
    }
    count as i32
}

fn factorial(n: usize) -> usize {
    if n == 0 {
        1
    } else {
        n * factorial(n - 1)
    }
}