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::num;
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(|e| e.parse().ok().unwrap()).collect()
}

fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
  (0..n).map(|_| read_vec()).collect()
}

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;
}

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

  let mut permutations: Vec<usize> = (0..n).map(|i| i+1).collect();
  let mut count = 0;
  let mut a= 0;
  let mut b= 0;
  loop {
    count += 1;
    if (0..n).all(|i| p[i] == permutations[i]) {
      a = count;
    }
    if (0..n).all(|i| q[i] == permutations[i]) {
      b = count;
    }
    if !permutations.next_permutation() { break; }
  }

  println!("{}", std::cmp::max(a, b) - std::cmp::min(a, b));

}