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 (1.42.0) (judge verdict: Accepted)

use std::io::Read;
use std::ops::Sub;

fn order(p: &[i32], i: usize, m: u32, o: usize) -> usize {
    let n = p.len();
    if i == n {
        o
    } else {
        let usable_digits = n - m.count_ones() as usize;
        let digits_to_p = (0..p[i]).filter(|&j| (m >> j) & 1 == 0).count();
        order(
            p,
            i + 1,
            m | (1 << (p[i] - 1)),
            o * usable_digits + digits_to_p,
        )
    }
}

fn abs_sub<T: PartialOrd + Sub>(a: T, b: T) -> T::Output {
    if a >= b {
        a - b
    } else {
        b - a
    }
}

fn solve(_n: usize, p: &[i32], q: &[i32]) -> usize {
    abs_sub(order(p, 0, 0, 0), order(q, 0, 0, 0))
}

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();
    let p = (0..n)
        .map(|_| iter.next().unwrap().parse().unwrap())
        .collect::<Vec<_>>();
    let q = (0..n)
        .map(|_| iter.next().unwrap().parse().unwrap())
        .collect::<Vec<_>>();
    let r = solve(n, &p, &q);
    println!("{}", r);
}