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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
    private int N;
    private int[] A, B;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        A = sc.IntArray();
        B = sc.IntArray();

        /*
         * 偶数番目
         * 裏返すと奇数番目になる
         *
         * 奇数番目
         * 裏返すと偶数番目になる
         *
         * 2^18全探索
         * 距離が合ってるか
         *
         * 転倒数
         */

        int l = 1 << N;
        int ans = int.MaxValue;
        for (int i = 0; i < l; i++)
        {
            int o;
            if (C(i, out o))
            {
                ans = Math.Min(o, ans);
            }
        }

        Console.WriteLine(ans == int.MaxValue ? -1 : ans);
    }

    private bool C(int l, out int cnt)
    {
        // false A面 true B面
        var b = new bool[N];
        int c = 0;
        cnt = 0;
        for (int i = 0; i < N; i++)
        {
            b[i] = (l & (1 << i)) > 0;
            if (b[i])
            {
                c++;
            }
        }

        if (c % 2 != 0)
        {
            return false;
        }

        var ar = new int[N];
        for (int i = 0; i < N; i++)
        {
            if (b[i])
            {
                ar[i] = B[i];
            }
            else
            {
                ar[i] = A[i];
            }
        }

        Array.Sort(ar);
        //Console.WriteLine(string.Join(" ",ar));
        bool[] flag = new bool[N];
        for (int i = 0; i < N; i++)
        {
            bool f = false;
            for (int j = 0; j < N; j++)
            {
                int n = b[j] ? B[i] : A[i];
                if (n == ar[j] && i % 2 == (j + (b[j] ? 1 : 0)) % 2 && !flag[j])
                {
                    cnt += Math.Abs(j - i);

                    f = true;
                    flag[j] = true;
                    break;
                }
            }

            if (!f)
            {
                return false;
            }
        }

        cnt /= 2;
        return true;
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

class HashMap : Dictionary<int, HashSet<int>>
{
    new public HashSet<int> this[int i]
    {
        get
        {
            HashSet<int> v;
            return TryGetValue(i, out v) ? v : base[i] = new HashSet<int>();
        }
        set { base[i] = value; }
    }
}

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}