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)

//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

// 裏の数は偶数
// 3回のコストでi, i + 2 swap
// AaBb と書くとリバース
// 裏表決め打ちしたときにソートできるか、最小値達成できるか
// コストの計算わからん
// 値が小さいのはなんか役に立たないか
// 対角線のswap

fn run() {
    input! {
        n: usize,
        a: [usize1; n],
        b: [usize1; n],
    }
    let inf = std::u16::MAX;
    let mut dp = vec![vec![inf; 1 << n]; 50];
    dp[0][0] = 0;
    for s in 0..50 {
        for i in 0..(1usize << n) {
            if dp[s][i] == inf {
                continue;
            }
            let v = dp[s][i];
            let c = i.count_ones() as usize & 1;
            let mut cost = 0;
            for j in 0..n {
                if (i >> j) & 1 == 1 {
                    continue;
                }
                cost += 1;
                let x = if c ^ (j & 1) == 0 {a[j]} else {b[j]};
                if x < s {
                    continue;
                }
                dp[x][i | (1 << j)] = std::cmp::min(dp[x][i | (1 << j)], v + cost - 1);
            }
        }
    }
    let mut ans = inf;
    for i in 0..n {
        ans = std::cmp::min(ans, dp[i][(1 << n) - 1]);
    }
    if ans == inf {
        println!("-1");
    } else {
        println!("{}", ans);
    }
}

fn main() {
    run();
}