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:
// D - Swap and Flip
#include <bits/stdc++.h>

using namespace std;

#define	INF	1000000000

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);

	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>(n, INF));
	dp[0][0] = 0;
	for (int k = 0; k < (1 << n); k++)
		for (int l = 0; l < n; l++) {
			if ((k & (1 << l)) == 0) continue;
			int p = k ^ (1 << l);
			int c = __builtin_popcount(p);
			int x = abs(l - c) % 2 == 0 ? a[l] : b[l];
			int d = __builtin_popcount((~k) & ((1 << l) - 1));
			if (p == 0) {
				dp[k][l] = d;
			} else {
				for (int i = 0; i < n; i++) {
					if ((p & (1 << i)) == 0) continue;
					int y = abs(i - c + 1) % 2 == 0 ? a[i] : b[i];
					if (y <= x) dp[k][l] = min(dp[k][l], dp[p][i] + d);
				}
			}
		}
	int ans = INF;
	for (int i = 0; i < n; i++) ans = min(ans, dp[(1 << n) - 1][i]);

	cout << (ans < INF ? ans : -1) << endl;

	return 0;
}