Problem:
Score : 700 points

Problem Statement

We have N cards numbered 1, 2, ..., N.
Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side.
Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

Choose an integer i (1 \leq i \leq N - 1).
Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

1 \leq N \leq 18

1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If it is impossible to have a non-decreasing sequence, print -1.
If it is possible, print the minimum number of operations required to achieve it.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let n = read::<usize>();
    let a = read_vec::<usize>();
    let b = read_vec::<usize>();

    let mut ans = std::i64::MAX;
    'outer: for i in 0..1 << n {
        let mut evens = vec![];
        let mut odds = vec![];
        for j in 0..n {
            let mut oe = j % 2;
            let mut num;
            if i >> j & 1 == 1 {
                oe = 1 - oe;
                num = b[j];
            } else {
                num = a[j];
            }
            if oe == 0 {
                evens.push((num, j));
            } else {
                odds.push((num, j));
            }
        }

        if odds.len() != n / 2 {
            continue 'outer;
        }

        evens.sort();
        odds.sort();
        let mut numbers = vec![];
        for i in 0..n / 2 {
            numbers.push(evens[i].0);
            numbers.push(odds[i].0);
        }
        if n % 2 == 1 {
            numbers.push(evens[n / 2].0);
        }
        for i in 0..n - 1 {
            if numbers[i] > numbers[i + 1] {
                continue 'outer;
            }
        }

        let mut temp = 0;
        for j in 0..evens.len() {
            temp += (evens[j].1 as i64 - 2 * j as i64).abs();
        }
        for j in 0..odds.len() {
            temp += (odds[j].1 as i64 - 2 * j as i64 - 1).abs();
        }
        ans = min(ans, temp);
    }
    if ans == std::i64::MAX {
        println!("-1");
    } else {
        println!("0000");
        //println!("{}", ans / 2);
    }
}

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()
}