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)

// input! macro is published in the following URLs.
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
// https://github.com/tanakh/competitive-rs.git
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn next_permutation(v: &mut [i64]) -> bool {
    if v.len() < 2 {
        return false;
    }

    let mut i = v.len() - 1;
    while i > 0 && v[i - 1] >= v[i] {
        i -= 1;
    }

    if i == 0 {
        return false;
    }

    let mut j = v.len() - 1;
    while j >= i && v[j] <= v[i - 1] {
        j -= 1;
    }

    v.swap(j, i - 1);

    v[i..].reverse();

    true
}

fn main() {
    input! {
        n: usize,
        ps: [i64; n],
        qs: [i64; n],
    }
    let mut v = vec![];
    for i in 1..n + 1 {
        v.push(i as i64);
    }

    let mut a = 1i32;
    loop {
        if v == ps {
            break;
        }
        next_permutation(v.as_mut_slice());
        a += 1;
    }

    v.sort();
    let mut b = 1;
    loop {
        if v == qs {
            break;
        }
        next_permutation(v.as_mut_slice());
        b += 1;
    }
    let ans = (a - b).abs();
    println!("{}", ans);
}