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, unused_macros, dead_code)]

macro_rules! min {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| min!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = min!($($ys),*);
        if $x < t { $x } else { t }
    }}
}
macro_rules! max {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| max!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = max!($($ys),*);
        if $x > t { $x } else { t }
    }}
}


macro_rules! ewriteln {
    ($($args:expr),*) => { let _ = writeln!(&mut std::io::stderr(), $($args),*); };
}
macro_rules! trace {
    ($x:expr) => { ewriteln!(">>> {} = {:?}", stringify!($x), $x) };
    ($($xs:expr),*) => { trace!(($($xs),*)) }
}
macro_rules! put {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        if let Some(x) = it.next() { print!("{}", x); }
        for x in it { print!(" {}", x); }
        println!("");
    }};
    ($x:expr) => { println!("{}", $x) };
    ($x:expr, $($xs:expr),*) => { print!("{} ", $x); put!($($xs),*) }
}

const M: i64 = 1_000_000_007;

// n^m
struct PowerPermutation { n: usize, m: usize, ar: Vec<usize> }
impl PowerPermutation {
    fn new(n: usize, m: usize) -> PowerPermutation {
        let ar = vec![0; m];
        PowerPermutation { n: n, m: m, ar: ar }
    }
}
impl Iterator for PowerPermutation {
    type Item = Vec<usize>;
    fn next(&mut self) -> Option<Vec<usize>> {
        if self.ar[self.m-1] >= self.n { return None }
        let r = self.ar.clone();
        self.ar[0] += 1;
        for i in 0..self.m-1 {
            if self.ar[i] == self.n {
                self.ar[i] = 0;
                self.ar[i + 1] += 1;
            } else {
                break;
            }
        }
        return Some(r);
    }
}

fn main() {
    let mut sc = Scanner::new();
    let n: usize = sc.cin();
    let xs: Vec<usize> = sc.vec(n);
    let ys: Vec<usize> = sc.vec(n);

    let mut ans = n + 1;
    for p in PowerPermutation::new(2, n) {
        let mut xs = xs.clone();
        let mut ys = ys.clone();

        let mut zs: Vec<usize> =
            (0..n).map(|i| if p[i] == 0 { xs[i] } else { ys[i] }).collect();
        zs.sort();

        let mut flip = 0;
        for i in 0..n-1 {
            let mut par = true;
            for j in 0..n-i {
                let it = if par {
                    xs[i + j] == zs[i]
                } else {
                    ys[i + j] == zs[i]
                };
                par ^= true;
                if it {
                    for k in 0..j {
                        {
                            let i = i + k;
                            flip += 1;
                            let tmp = xs[i];
                            let tmp2 = ys[i];
                            xs[i] = ys[i+1];
                            ys[i] = xs[i+1];
                            xs[i+1] = tmp2;
                            ys[i+1] = tmp;
                        }
                    }
                    break
                }
            }
        }
        let mut ok = true;
        for i in 0..n-1 {
            if xs[i] > xs[i+1] {
                ok = false;
            }
        }
        if ok {
            ans = min!(ans, flip);
        }
    }

    if ans == n + 1 {
        put!(-1);
    } else {
        put!(ans);
    }


}

use std::io::{self, Write};
use std::str::FromStr;
use std::collections::VecDeque;

struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
impl Scanner {
    fn new() -> Self { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn cin<T: FromStr>(&mut self) -> T {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
        self.buffer.pop_front().unwrap().parse::<T>().ok().unwrap()
    }
    fn chars(&mut self) -> Vec<char> { self.cin::<String>().chars().collect() }
    fn vec<T: FromStr>(&mut self, n: usize) -> Vec<T> { (0..n).map(|_| self.cin()).collect() }
}