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 <bits/stdc++.h>
using namespace std;

signed main() {
	ios::sync_with_stdio(0); cin.tie(0);

	int n; cin >> n;
	vector<int> a(n), b(n);
	for (int &i: a) cin >> i;
	for (int &i: b) cin >> i;

	int ans = 998244353;
	for (int m = 0; m < (1 << n); m++) {
		if (__builtin_popcount(m) != n/2) continue;

		vector<pair<int, int>> e, o;
		for (int j = 0; j < n; j++) {
			if (m & (1 << j)) {
				if (j % 2) o.emplace_back(a[j], j);
				else o.emplace_back(b[j], j);
			} else {
				if (j % 2) e.emplace_back(b[j], j);
				else e.emplace_back(a[j], j);
			}
		} 
		sort(e.begin(), e.end());
		sort(o.begin(), o.end());

		vector<pair<int, int>> xd;
		for (int i = 0; i < e.size(); i++) {
			xd.push_back(e[i]);
			if (i < o.size()) xd.push_back(o[i]);
		}

		int tmp = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i+1; j < n; j++) {
				if (xd[i].first > xd[j].first) tmp = 998244353;

				tmp += (xd[i].second > xd[j].second);
			}
		} 
		ans = min(ans, tmp);
	}
	cout << (ans == 998244353 ? -1 : ans) << '\n';

	return 0;
}