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::{self, BufRead};

fn main() {
    let stdin = io::stdin();
    let mut lines = stdin.lock().lines();
    let n: i32 = lines.next().unwrap().unwrap().parse().unwrap();
    let p = lines.next().unwrap().unwrap();
    let p: Vec<i32> = p.split_whitespace().map(|s| s.parse().unwrap()).collect();
    let q = lines.next().unwrap().unwrap();
    let q: Vec<i32> = q.split_whitespace().map(|s| s.parse().unwrap()).collect();

    fn permutation(v: Vec<i32>) -> Vec<Vec<i32>> {
        if v.len() == 1 {
            return vec![v];
        }
        let mut result = Vec::new();
        for i in 0..v.len() {
            let mut new_v = v.clone();
            let first = new_v.remove(i);
            for mut p in permutation(new_v) {
                p.insert(0, first);
                result.push(p);
            }
        }
        result
    }

    let perm = permutation((1..n + 1).collect());

    let a = perm.iter().position(|v| *v == p).unwrap();
    let b = perm.iter().position(|v| *v == q).unwrap();

    let ans = (a as i32 - b as i32).abs();
    println!("{}", ans);
}