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 proconio::{fastout, input};

pub trait LexicalPermutation {
    /// Return  if the slice was permuted,  if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return  if the slice was permuted,  if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i - 1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j - 1] < self[i - 1] {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i - 1, j);

        true
    }
}

#[fastout]
fn main() {
    input! {
        n: usize,
        p: [usize; n],
        q: [usize; n]
    }

    let find_prm = |v: Vec<usize>| -> isize {
        // let mut prm = (1..=n).collect::<Vec<usize>>();
        let mut prm = (1..=n).collect::<Vec<usize>>();
        let mut prm_index = 0;
        loop {
            prm_index += 1;
            if v == prm {
                return prm_index;
            }

            if !prm.next_permutation() {
                panic!();
            }
        }
    };

    println!("{:?}", (find_prm(p) - find_prm(q)).abs());
}