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::Write;

#[allow(unused)]
macro_rules! debug {
    ($($format:tt)*) => (write!(std::io::stderr(), $($format)*).unwrap());
}
#[allow(unused)]
macro_rules! debugln {
    ($($format:tt)*) => (writeln!(std::io::stderr(), $($format)*).unwrap());
}

fn main() {
    let n: usize = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        let s = s.trim_right().to_owned();
        s.parse().unwrap()
    };

    let p: Vec<usize> = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        let s = s.trim_right().to_owned();
        s.split_whitespace().map(|x| x.parse().unwrap()).collect()
    };

    let q: Vec<usize> = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        let s = s.trim_right().to_owned();
        s.split_whitespace().map(|x| x.parse().unwrap()).collect()
    };

    let ord1 = calc_order(&p);
    let ord2 = calc_order(&q);
    if ord1 < ord2 {
        println!("{}", ord2 - ord1);
    } else {
        println!("{}", ord1 - ord2);
    }
}

fn calc_order(v: &Vec<usize>) -> usize {
    let n = v.len();

    let mut used = vec![false; n + 1];
    used[0] = true;

    let mut ret = 0;
    for i in 0..n {
        let a = used[1..v[i]].iter().filter(|x| **x == false).count();
        // debugln!("{}, {}", n, a);
        ret += a * factorial(n - i - 1);
        used[v[i]] = true;
    }

    ret + 1
}

fn factorial(n: usize) -> usize {
    let mut ret = 1;
    for i in 1..(n + 1) {
        ret *= i;
    }
    ret
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn factorial1() {
        assert_eq!(6, factorial(3));
    }

    #[test]
    fn factorial2() {
        assert_eq!(24, factorial(4));
    }

    #[test]
    fn calc_order1() {
        assert_eq!(5, calc_order(&vec![3, 1, 2]));
        assert_eq!(2, calc_order(&vec![1, 3, 2]));
    }
}