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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int n,ans = 20;
int a[60],b[60];
void dfs(int pos,int t)
{
	if(pos == n)
	{
		bool flag = 1;
		for(int i = 1; i < n; i++)
			if(a[i+1] < a[i])
				flag = 0;
		if(!flag)	return;
		ans = min(ans, t);
		return;
	}
	dfs(pos + 1, t);
	swap(a[pos], b[pos+1]);
	swap(a[pos+1], b[pos]);
	dfs(pos + 1,  t+1);
	swap(a[pos+1], b[pos]);
	swap(a[pos], b[pos+1]);
}
 
int main()
{
	scanf("%d",&n);
	for(int i = 1; i <= n; i++)	scanf("%d",&a[i]);
	for(int i = 1; i <= n; i++)	scanf("%d",&b[i]);
	dfs(1, 0);
	if(ans == 20)	printf("-1");
	else	printf("%d",ans);
	return 0;
}