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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.NoSuchElementException;
import java.util.StringTokenizer;

import static java.lang.Math.min;

public class Main {

    public static void main(String[] args) throws IOException{
        new Main().solve();
    }

    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer st;

    int readInt() throws IOException {
        if (canRead()) {
            return Integer.parseInt(st.nextToken());
        }
        throw new NoSuchElementException();
    }
    long readLong() throws IOException {
        if (canRead()) {
            return Long.parseLong(st.nextToken());
        }
        throw new NoSuchElementException();
    }

    char readChar() throws IOException {
        if (canRead()) {
            return st.nextToken().charAt(0);
        }
        throw new NoSuchElementException();
    }

    boolean canRead() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            String s = in.readLine();
            if (s != null) {
                st = new StringTokenizer(s, " ");
            } else {
                return false;
            }
        }
        return true;
    }

    int MAXN = 20;
    int INF = 1000000007;
    int N;
    int[] A = new int[MAXN], B = new int[MAXN];
    private void solve() throws IOException {
        N = readInt();
        for (int i = 0; i < N; i++) {
            A[i] = readInt();
        }
        for (int i = 0; i < N; i++) {
            B[i] = readInt();
        }
        dosolve();
        int ans = INF;
        for (int i = 0; i < N; i++) {
            ans = min(ans, dp[(1<<N)-1][i]);
        }
        out.println(ans == INF ? -1 : ans);
        out.flush();
    }

    int[][] dp = new int[1<<MAXN][MAXN];
    private void dosolve() {
        for (int S = 0; S < (1 << N); S++) {
            Arrays.fill(dp[S], INF);
        }
        for (int i = 0; i < N; i++) {
            dp[1<<i][i] = 0;
        }
        for (int S = 0; S < (1 << N); S++) {
            for (int i = 0; i < N; i++) {
                if ((S>>i&1) == 1) {
                    int c = Integer.bitCount(S);
                    int vi = (c-1&1) == (i&1) ? A[i] : B[i];
                    int cost = c;
                    for (int j = 0; j < N; j++) {
                        if ((S>>j&1) == 1) {
                            cost--;
                            continue;
                        }
                        int vj = (c&1) == (j&1)? A[j] : B[j];
                        if (vj >= vi) {
                            dp[S|1<<j][j] = min(dp[S|1<<j][j], dp[S][i] + cost);
                        }
                    }
                }
            }
        }
    }


}