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.util.*;
public class Main {
	
	public static void main(String[] args) {
		int INF = Integer.MAX_VALUE/2;
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] A = new int[N];
		Arrays.setAll(A, i -> sc.nextInt());
		int[] B = new int[N];
		Arrays.setAll(B, i -> sc.nextInt());
		int ans = INF;
		int[][] dp = new int[1<<N][51];
		for(int i = 0; i < 1<<N; i++)
			Arrays.fill(dp[i], INF);
		dp[0][0] = 0;
		for(int S = 0; S < 1<<N; S++) {
			int j = Integer.bitCount(S);
			int c = 0;
			for(int i = 0; i < N; i++) {
				if((S & 1<<i) == 0) {
					int k = (i - j) % 2 == 0 ? A[i] : B[i];
					for(int l = 0; l <= k; l++)
						dp[S|1<<i][k] = Math.min(dp[S|1<<i][k], dp[S][l]+c);
					c++;
				}
			}
		}
		for(int i : dp[(1<<N)-1])
			ans = Math.min(ans, i);
		System.out.println(ans == INF ? -1 : ans);
	}

}
