Problem: 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.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
using namespace std;	

int a[29], b[29], r[29], d[29];
vector<pair<int, int> > av, bv;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n; cin >> n;
	for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 0; i < n; i++) {
		cin >> b[i];
		if (i % 2) swap(a[i], b[i]);
	}
	int ans = 1000000000;
	for (int i = 0; i < (1 << n); i++) {
		int t = i, c = 0;
		while (t) {
			t -= (t & (-t)); c++;
		}
		if (c != (n + 1) / 2) continue;
		av.clear(); bv.clear();
		for (int j = 0; j < n; j++) {
			if (i & (1 << j)) av.push_back(make_pair(a[j], j));
			else bv.push_back(make_pair(b[j], j));
		}
		sort(av.begin(), av.end());
		sort(bv.begin(), bv.end());
		bool f = true;
		for (int j = 0; j < n; j++) {
			if (j % 2) { r[j] = bv[j / 2].second; d[j] = bv[j / 2].first; }
			else { r[j] = av[j / 2].second; d[j] = av[j / 2].first; }
			if (j > 0 && d[j] < d[j - 1]) {
				f = false;
				break;
			}
		}
		if (!f) continue;
		int s = 0;
		for (int j = 0; j < n; j++)
			for (int k = j + 1; k < n; k++)
				if (r[j] > r[k])
					s++;
		ans = min(ans, s);
	}
	if (ans == 1000000000) cout << -1 << '\n';
	else cout << ans << '\n';
	return 0;
}