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)

use std::io;
use std::str::FromStr;

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

fn input_vec() -> Vec<u64> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}

fn factorial(n: u64) -> u64 {
    let mut res = 1;
    let mut i = n;
    while i > 0 {
        res *= i;
        i -= 1;
    }
    res
}

fn order(target_vec: &Vec<u64>, num_list: &mut Vec<u64>) -> u64 {
    // println!("{:?}", target_vec);
    let mut sum = 0;
    for i in 0..target_vec.len() {
        let mut idx = 0;
        for j in 0..num_list.len() {
            if target_vec[i] == num_list[j] {
                idx = j;
                break;
            }
        }
        num_list.remove(idx);
        sum += idx as u64 * factorial((target_vec.len()- 1 - i) as u64);
    }
    sum
}

fn main() {
    let n: u64 = input();
    let p_vec = input_vec();
    let q_vec = input_vec();
    let mut num_list: Vec<u64> = Vec::new();
    for i in 1..(n+1) {
        num_list.push(i);
    }
    let p = order(&p_vec, &mut num_list);
    let mut num_list: Vec<u64> = Vec::new();
    for i in 1..(n+1) {
        num_list.push(i);
    }
    let q = order(&q_vec, &mut num_list);
    if p < q {
        println!("{}", q - p);
    } else {
        println!("{}", p - q);
    }

}