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
import java.io.*;
import java.util.*;

class Main {

	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();

		int[][] a = new int[n][2];
		for (int i = 0; i < n; i++) {
			a[i][0] = sc.nextInt();
		}
		for (int i = 0; i < n; i++) {
			a[i][1] = sc.nextInt();
		}

		int dp[][] = new int[1 << n][51];
		for (int i = 0; i < 1 << n; i++) {
			Arrays.fill(dp[i], Integer.MAX_VALUE / 3);
		}

		int[][] swap = new int[n][n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				swap[i][j] = (i % 2) == (j % 2) ? a[j][0] : a[j][1];
			}
		}

		dp[0][0] = 0;
		for (int mask = 0; mask < (1 << n); mask++) {
			int x = Integer.bitCount(mask);

			for (int val = 0; val < 51; val++) {
				if (dp[mask][val] == Integer.MAX_VALUE / 3) continue;

				int cost = 0;
				for (int i = 0; i < n; i++) {
					if ((mask & (1 << i)) > 0) continue;
					if (swap[x][i] >= val) {
						dp[mask | (1 << i)][swap[x][i]] = Math.min(
								dp[mask | (1 << i)][swap[x][i]],
								dp[mask][val] + cost);
					}
					cost++;
				}
			}
		}

		int answer = Integer.MAX_VALUE;
		for (int i = 0; i < 51; i++) {
			answer = Math.min(answer, dp[(1 << n) - 1][i]);
		}

		if (answer == Integer.MAX_VALUE / 3) {
			System.out.println(-1);
		} else {
			System.out.println(answer);
		}
	}
}
