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 (1.42.0) (judge verdict: Wrong Answer)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let a: Vec<usize> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    let b: Vec<usize> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();

    // a ... 偶数回スワップして表
    // b ... 奇数回スワップして表
    const INF: i32 = 1 << 30;
    let mut ans = INF;
    for bit in 0..1 << n {
        let mut tmp = Vec::new();
        for i in 0..n {
            if bit >> i & 1 == 0 {
                tmp.push((a[i], i, 0));
            } else {
                tmp.push((b[i], i, 1));
            }
        }
        tmp.sort();

        let mut ok = true;
        let mut cnt = 0;
        for i in 0..n {
            let v = tmp[i].0;
            let j = tmp[i].1;
            let k = tmp[i].2;
            let diff = (i as i32 - j as i32).abs();
            if diff % 2 != k {
                ok = false;
            }
            cnt += diff;
        }
        if ok {
            ans = min(ans, cnt / 2);
        }
    }
    if ans == INF {
        println!("-1")
    } else {
        println!("{}", ans);
    }
}