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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include<math.h>
int fun(int a[], int n)
{
	if (n == 1)
		return 1;
	if (n == 2)
		return a[n - 1] <= a[n - 2];
	return fun(a, n - 1) && (a[n - 1] <= a[n - 2]);
}
void huan(int *a,int *b)
{
	int *t=0;
	*t = *a; *a = *b; *b = *t;
}
int main()
{
	int n, a[19], b[19];
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
	}
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &b[i]);
	}
	if (fun(a, n) == 0)
	{
		printf("0");
		return 0;
	}
	int j = 1,t;
	while (j <= n - 2)
	{
		t = a[j]; a[j] = b[j]; b[j] = t;
		t = a[j-1]; a[j-1] = b[j-1]; b[j-1] = t;
		t = a[j]; a[j] = a[j-1]; a[j-1] = t;
		if (fun(a, n) == 1) { j++; continue; }
		else {
			printf("%d", j); break;
		}
	}
	if (j > n - 2)printf("-1");
}