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 find_index(ps: &Vec<usize>, n: usize) -> usize {
  let mut digits: Vec<usize> = (1..(n+1)).collect();
  let mut total: usize = 0;
  for i in 0..n {
    if let Some(j) = digits.iter().position(|&x| x == ps[i]) {
      let x =  j * (1..(n-i)).product::<usize>();
      total += x;
      digits.remove(j);
    }
  }
  total
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let line = lines.next().unwrap().unwrap();
  let n: usize = line.parse().unwrap();
  let line = lines.next().unwrap().unwrap();
  let ps: Vec<usize> = line.split(' ').map(|x| x.parse().unwrap()).collect();
  let line = lines.next().unwrap().unwrap();
  let qs: Vec<usize> = line.split(' ').map(|x| x.parse().unwrap()).collect();
  let x = find_index(&ps, n);
  let y = find_index(&qs, n);
  let ans = if x < y { y - x } else { x - y };
  println!("{}", ans);
}