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: Accepted)

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define x first
#define y second

using namespace std;

typedef long long ll;

int a[20], b[20];

int main()
{
 	//freopen("input.txt", "r", stdin);
 	//freopen("output.txt", "w", stdout);
 	ios_base::sync_with_stdio(0); cin.tie(0);
 	int n;
 	cin >> n;
 	for (int i = 1; i <= n; i++) cin >> a[i];
 	for (int i = 1; i <= n; i++) cin >> b[i];

 	int ans = -1;
 	for (int mask = 0; mask < (1 << n); mask++)
 	{
 	 	vector <pair <int, int> > spis1, spis2, spis;
 	 	int odd = 0, even = 0;
 	 	for (int i = 1; i <= n; i++)
 	 		if (((1 << (i - 1)) & mask) != 0) odd++;
 	 		else even++;
 	   	if (odd != (n + 1) / 2 || even != n / 2) continue;
 	   	
 	   	for (int i = 1; i <= n; i++)
 	 		if (((1 << (i - 1)) & mask) != 0)
 	 		{
 	 		 	if (i % 2 == 1) spis1.pb(mp(a[i], i));
 	 		 	else spis1.pb(mp(b[i], i));
 	 		}
 	 		else
 	 		{
 	 		 	if (i % 2 == 0) spis2.pb(mp(a[i], i));
 	 		 	else spis2.pb(mp(b[i], i));
 	 		}
 	   	sort(spis1.begin(), spis1.end());
 	   	sort(spis2.begin(), spis2.end());
 	   	
 	   	int uk1 = 0, uk2 = 0;
 	   	while (uk1 < (int) spis1.size() || uk2 < (int) spis2.size())
 	   	{
 	   		if (uk1 < (int) spis1.size()) spis.pb(spis1[uk1]);
 	   		if (uk2 < (int) spis2.size()) spis.pb(spis2[uk2]);
 	   		uk1++, uk2++; 	
 	   	}
        
        bool fl = true;
 	   	for (int i = 0; i < n - 1; i++) if (spis[i].x > spis[i + 1].x) fl = false;
 	   	if (!fl) continue;
 	   	
 	   	int tek = 0;
 	   	for (int i = 0; i < n; i++)
 	   		for (int j = i + 1; j < n; j++)
 	   			if (spis[i].y > spis[j].y) tek++;
 	   	if (ans == -1) ans = tek;
 	   	else ans = min(ans, tek);
 	}
 	cout << ans << '\n';
 	return 0;
}