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)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|s| s.parse().ok().unwrap())
        .collect()
}

// fn dfs(unused: &[usize], used: &[usize], target: &[usize], count: usize) -> usize {
//     if target == used {
//         return count;
//     }
//     for i in 0..unused.len() {
//         dfs(unused.iter().enumerate().filter(|&(j,v)| j != i).map(|&(j,v) v).collect(), )
//     }

// }

fn next_permutation<T: Ord>(v: &mut [T]) -> bool {
    let mut i = v.len() - 1;
    while i > 0 && v[i-1] > v[i] {
        i -= 1;
    }

    if i == 0 {
        v.reverse();
        return false;
    }
    i -= 1;
    let mut j = v.len() - 1;
    while j > i && v[i] > v[j] {
        j -= 1;
    }
    v.swap(i,j);
    v[i + 1..].reverse();
    true
}

fn main() {
    let n = read::<usize>();
    let p = read_vec::<usize>();
    let q = read_vec::<usize>();

    let mut v = vec![];
    for i in 1..n+1 {
        v.push(i);
    }
    let mut p_count: i64 = 0;
    let mut q_count: i64 = 0;

    let mut count: i64 = 1;
    loop {
        if p == v {
            p_count = count;
        }
        if q == v {
            q_count = count;
        }

        count += 1;
        if !next_permutation(&mut v) {
            break;
        }
    }

    println!("{}", (p_count- q_count).abs());

}