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(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let A: Vec<usize> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };
    let B: Vec<usize> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };

    const INF: usize = 1_000_000_000;
    let res = (0..(1 << N))
        .map(|p| {
            let mut a = vec![];
            let mut b = vec![];
            for i in 0..N {
                if (p >> i) & 1 == 1 {
                    a.push((A[i], i));
                } else {
                    b.push((B[i], i));
                }
            }
            a.sort();
            b.sort();
            // println!("a={:?}", a);
            // println!("b={:?}", b);
            let mut X = vec![];
            for i in 0..N {
                X.push(i);
            }
            let mut x = 0;
            let mut cost = 0;
            for i in 0..N {
                // println!("{:?}", X);
                let j = (0..a.len()).find(|&j| a[j].1 % 2 == i % 2).unwrap_or(INF);
                let k = (0..b.len()).find(|&k| b[k].1 % 2 != i % 2).unwrap_or(INF);
                if j < INF && x <= a[j].0 && (k == INF || a[j].0 < b[k].0) {
                    x = a[j].0;
                    let l = (0..N).find(|&l| X[l] == a[j].1).unwrap();
                    // println!("{}", l);
                    cost += l - i;
                    for r in (i..l).rev() {
                        let tmp = X[r + 1];
                        X[r + 1] = X[r];
                        X[r] = tmp;
                    }
                    a.remove(j);
                } else if k < INF && x <= b[k].0 && (j == INF || b[k].0 <= a[j].0) {
                    x = b[k].0;
                    let l = (0..N).find(|&l| X[l] == b[k].1).unwrap();
                    cost += l - i;
                    for r in (i..l).rev() {
                        let tmp = X[r + 1];
                        X[r + 1] = X[r];
                        X[r] = tmp;
                    }
                    b.remove(k);
                } else {
                    cost = INF;
                    break;
                }
            }
            cost
        })
        .min()
        .unwrap();
    let ans = if res == INF { -1 } else { res as i64 };

    println!("{}", ans);
}