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.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		int n = int.Parse(Console.ReadLine());
		int[] numsA = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
    int[] numsB = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
    int[,] numsSortBefore = new int[n,3];//数値と表裏必要と現在
    int changeCount = 0;
    int answer = 99999;
    bool canChange = true;

    for(long i = 0; i < (1<<n); i++)//２進数で全通り試す
    {
      changeCount = 0;
      canChange = true;
      for(int j = 0; j < n; j++)//00001などに1があるか右側から見る
      {
        if((i&(1<<j)) != 0)
        {
          numsSortBefore[j,0] = numsA[j];
          numsSortBefore[j,1] = 0;
          numsSortBefore[j,2] = 0;
        }
        else//0の場合の処理
        {
          numsSortBefore[j,0] = numsB[j];
          numsSortBefore[j,1] = 1;
          numsSortBefore[j,2] = 0;
        }
        //Console.WriteLine(i+"a"+j+" "+numsSortBefore[j,0]);
      }
  
      for(int j = n-1; j > 0; j--)
      {
        for(int k = j; k > 0; k--)
        {
          if(numsSortBefore[k-1,0] > numsSortBefore[k,0])
          {
            int swapMemo = numsSortBefore[k-1,0];
            numsSortBefore[k-1,0] = numsSortBefore[k,0];
            numsSortBefore[k,0] = swapMemo;

            numsSortBefore[k-1,2] = 1-numsSortBefore[k-1,2];
            numsSortBefore[k,2] = 1-numsSortBefore[k,2];

            changeCount++;
          }
        }
      }

      for(int j = 0; j < n; j++)
      {
        //Console.WriteLine(i+" "+numsSortBefore[j,1]+" "+numsSortBefore[j,2]);
        if(numsSortBefore[j,1] != numsSortBefore[j,2])
        {
          if(j == n-1)
          {
            canChange = false;
            break;
          }else if((numsSortBefore[j+1,1] != numsSortBefore[j+1,2]) && (numsSortBefore[j,0] == numsSortBefore[j+1,0]))
          {
            changeCount++;
            j++;
          }else
          {
            canChange = false;
            break;
          }
        }
      }

      //Console.WriteLine(i+" "+changeCount+" "+canChange);
      if(canChange) answer = Math.Min(answer,changeCount);
      
    }

    if(answer == 99999) answer = -1;
		Console.WriteLine(answer);
	}
}