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>
#define Int int64_t

using namespace std;

int main() {
	const int INF = 1e9;
	int N;
	cin >> N;
	vector<int> a(N), b(N);
	for (int i = 0; i < N; ++i) { cin >> a[i]; }
	for (int i = 0; i < N; ++i) { cin >> b[i]; }

	vector<vector<int>> dp((1<<N), vector<int>(55, INF));
	dp[0][0] = 0;
	for (int i = 0; i < (1<<N); ++i) {
		int n = __builtin_popcount(i);
		int cost = 0;
		for (int j = 0; j < N; ++j) {
			if ((i>>j) & 1) { continue; }

			int ii = i | (1<<j);
			int val = (abs(n - j) % 2 ? b[j] : a[j]);
			for (int k = 0; k < 55; ++k) {
				if (k > val || dp[i][k] == INF) { continue; }
				dp[ii][val] = min(dp[ii][val], dp[i][k] + cost);
			}
			++cost;
		}
	}
	int ans = INF;
	for (int i = 0; i < 55; ++i) { ans = min(ans, dp[(1<<N) - 1][i]); }
	cout << (ans < INF ? ans : -1) << endl;

	return 0;
}