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<tuple>

using namespace std;

typedef long long ll;
typedef tuple<int, int, bool> T3;//値、何番目、表裏

int N;
int A[50], B[50];

const int INF = 100000000;

int dp[1 << 18][50 + 5];

int main() {
	cin >> N;
	for (int i = 0; i < N; i++)cin >> A[i];
	for (int i = 0; i < N; i++)cin >> B[i];

	for (int i = 0; i < (1 << N); i++)for (int j = 0; j <= 50; j++)dp[i][j] = INF;

	dp[0][0] = 0;
	for (int bit = 0; bit < (1 << N); bit++) {
		int nowpos = 0;
		for (int i = 0; i < N; i++)
			if ((bit >> i) & 1)nowpos++;
		for (int i = 0; i <= 50; i++) {
			if (dp[bit][i] == INF)continue;
			int cost = 0;
			for (int j = 0; j < N; j++) {
				if ((bit >> j) & 1)continue;
				int target = (j - nowpos + 100) % 2 == 0 ? A[j] : B[j];
				if (target >= i) {
					dp[bit | (1 << j)][target] = min(dp[bit | (1 << j)][target], dp[bit][i] + cost);
				}
				cost++;
			}
		}
	}
	int ans = INF;
	for (int i = 0; i <= 50; i++) {
		ans = min(ans, dp[(1 << N) - 1][i]);
	}
	if (ans == INF)cout << -1 << endl;
	else cout << ans << endl;
	return 0;
}