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: Accepted)

use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    let n: usize = parse!(it);
    let a: Vec<usize> = parse!(it, n);
    let b: Vec<usize> = parse!(it, n);
    let mut dp = vec![vec![None; n]; 1 << n];
    let diff = |x, y| if x > y { x - y } else { y - x };
    let v = |i: usize, to: usize| {
        if diff(i, to) % 2 == 1 {
            b[i]
        } else {
            a[i]
        }
    };
    for i in 0..n {
        dp[1 << i][i] = Some(0);
    }
    for i in 1usize..(1 << n) {
        let r = i.count_ones() as usize;
        for j in 0..n {
            if let Some(x) = dp[i][j] {
                for k in 0..n {
                    if (i >> k) & 1 == 1 || j == k || v(j, r - 1) > v(k, r) {
                        continue;
                    }
                    let largerk = (i & (!((1 << k) - 1))).count_ones() as usize;
                    dp[i | (1 << k)][k] = match dp[i | 1 << k][k] {
                        Some(y) => Some(std::cmp::min(y, x + largerk)),
                        None => Some(x + largerk),
                    }
                }
            }
        }
    }
    let mut ret = None;
    for i in 0..n {
        ret = match dp[(1 << n) - 1][i] {
            None => ret,
            Some(x) => match ret {
                None => Some(x),
                Some(ret) => Some(std::cmp::min(x, ret)),
            },
        }
    }
    match ret {
        None => println!("{}", -1),
        Some(x) => println!("{}", x),
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
3
3 4 3
3 2 3
        ";
        solve(s);
    }
}