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)

#[allow(unused_macros, dead_code)]
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, dead_code)]
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, dead_code)]
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<_>>()
    };

    ($next:expr, chars) => {
        read_value!($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")
    };
}

#[allow(dead_code)]
struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

#[allow(dead_code)]
impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut p = vec![0; n];
        for i in 0..n {
            p[i] = i;
        }
        return UnionFind {
            parent: p,
            rank: vec![0; n],
            size: vec![1; n],
        };
    }

    fn find(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let p = self.parent[x];
            let pr = self.find(p);
            self.parent[x] = pr;
            pr
        }
    }

    fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    fn unite(&mut self, a: usize, b: usize) {
        let a_root = self.find(a);
        let b_root = self.find(b);
        if self.rank[a_root] > self.rank[b_root] {
            self.parent[b_root] = a_root;
            self.size[a_root] += self.size[b_root];
        } else {
            self.parent[a_root] = b_root;
            self.size[b_root] += self.size[a_root];
            if self.rank[a_root] == self.rank[b_root] {
                self.rank[b_root] += 1;
            }
        }
    }

    fn get_size(&mut self, x: usize) -> usize {
        let root = self.find(x);
        self.size[root]
    }
}

const MOD_P: usize = 1000000007;

#[allow(dead_code)]
// fact(n) = n! mod p
fn fact(n: usize) -> usize {
    let mut acc = 1;
    for i in 1..n + 1 {
        acc = acc * i % MOD_P;
    }
    acc
}

#[allow(dead_code)]
fn mod_pow(b: usize, mut e: usize) -> usize {
    let mut base = b;
    let mut acc = 1;

    while e > 1 {
        if e % 2 == 1 {
            acc = acc * base % MOD_P;
        }
        e /= 2;
        base = base * base % MOD_P;
    }

    if e == 1 {
        acc = acc * base % MOD_P;
    }

    acc
}

#[allow(dead_code)]
fn comb(n: usize, r: usize) -> usize {
    // nCr = n! / (r! (n-r)!) = n! (r!)^(p-2) ((n-r)!)^(p-2)
    fact(n) * mod_pow(fact(r), MOD_P - 2) % MOD_P * mod_pow(fact(n - r), MOD_P - 2) % MOD_P
}

#[allow(dead_code)]
fn getline() -> String {
    let mut __ret = String::new();
    std::io::stdin().read_line(&mut __ret).ok();
    return __ret;
}

// copy from https://github.com/bluss/permutohedron/blob/master/src/lexical.rs
pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` 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
    }
}

#[test]
fn lexical() {
    let mut data = [1, 2, 3];
    data.next_permutation();
    assert_eq!(&data, &[1, 3, 2]);
    data.next_permutation();
    assert_eq!(&data, &[2, 1, 3]);
    data.prev_permutation();
    assert_eq!(&data, &[1, 3, 2]);
    data.prev_permutation();
    assert_eq!(&data, &[1, 2, 3]);
    assert!(!data.prev_permutation());
    let mut c = 0;
    while data.next_permutation() {
        c += 1;
    }
    assert_eq!(c, 5);
}
fn main() {
    input! {
        n: usize,
        p: [usize;n],
        q: [usize;n],
    }

    let mut x: Vec<usize> = (0..n).map(|i| i + 1).collect();
    let mut c = 0;
    let mut a = 0;
    let mut b = 0;
    loop {
        if (0..n).all(|i| p[i] == x[i]) {
            a = c;
        }
        if (0..n).all(|i| q[i] == x[i]) {
            b = c;
        }
        if !x.next_permutation() {
            break;
        }
        c += 1;
    }
    println!("{}", std::cmp::max(a, b) - std::cmp::min(a, b));
}