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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        var sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        var a = new int[n];
        Arrays.setAll(a, $ -> Integer.parseInt(sc.next()));
        var b = new int[n];
        Arrays.setAll(b, $ -> Integer.parseInt(sc.next()));
        sc.close();

        var x = new int[n];
        int inf = 1000000000;
        int min = inf;
        for (int k = 0; k < 1 << n; k++) {
            for (int i = 0; i < n; i++) {
                if ((k & (1 << i)) == 0) { // Front
                    x[i] = a[i];
                } else {                   // Back
                    x[i] = ~b[i];
                }
            }
            // Normal Bubblesort. 
            // The variable c means the inversion number (= the number of swap-and-flip operations) of the array x.
            int c = bubblesort(x, n);
            // Adjusting Front/Back.
            for (int i = 0; i < n - 1; i++) {
                if (x[i] < 0) {
                    // ~2, 2, 2, ~2 => ~2, ~2, ~2, ~2
                    if (x[i + 1] >= 0 && i + 2 < n) {
                        swapAndFlip(x, i + 1, i + 2);
                        c++;
                    }
                    // ~2, ~2, ~2, ~2 => 2, 2, ~2, ~2
                    swapAndFlip(x, i, i + 1);
                    c++;
                }
            }
            // Making sure the array x is correctly sorted
            if (allPositive(x, n) && isAscending(x, n)) {
                min = Math.min(min, c);
            }
        }
        System.out.println(min == inf ? -1 : min);
    }

    public static int bubblesort(int[] x, int n) {
        int c = 0;
        for (int i = 0; i < n - 1; i++) {
            for (int j = n - 1; j > i; j--) {
                int u = x[j - 1] < 0 ? ~x[j - 1] : x[j - 1];
                int v = x[j]     < 0 ? ~x[j]     : x[j]    ;
                if (u > v) {
                    swapAndFlip(x, j - 1, j);
                    c++;
                }
            }
        }
        return c;
    }

    public static void swapAndFlip(int[] x, int i, int j) {
        int tmp = x[i];
        x[i] = ~x[j];
        x[j] = ~tmp;
    }

    public static boolean allPositive(int[] x, int n) {
        for (int i = 0; i < n; i++) {
            if (x[i] < 0) {
                return false;
            }
        }
        return true;
    }

    public static boolean isAscending(int[] x, int n) {
        for (int i = 0; i < n - 1; i++) {
            if (x[i] > x[i + 1]) {
                return false;
            }
        }
        return true;
    }
}
