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)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
    read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
    match n {
        0 => a,
        _ => {
            a.push(read::<T>());
            read_n_logic(n - 1, a)
        }
    }
}

use std::cmp::{max, min};
fn main() {
    let n: usize = read();
    let p: Vec<i64> = read_n(n);
    let q: Vec<i64> = read_n(n);

    let mut total = 1;

    for i in 1..n + 1 {
        total *= i;
    }

    let mut total2 = 1;

    for i in 1..n + 1 {
        total2 *= i;
    }
    let mut c = vec![];
    for i in 1..n + 1 {
        c.push(i as i64);
    }
    let mut ans = 0;

    for i in p {
        let idx = c.iter().position(|&r| r == i).unwrap();
        ans += total / c.len() * (idx);
        total /= c.len();
        c.remove(idx);
    }
    let mut d = vec![];
    for i in 1..n + 1 {
        d.push(i as i64);
    }
    let mut ans2 = 0;

    for i in q {
        let idx = d.iter().position(|&r| r == i).unwrap();
        ans2 += total2 / d.len() * (idx);
        total2 /= d.len();
        d.remove(idx);
    }
    println!("{}", max(ans, ans2) - min(ans, ans2));
}