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 <vector>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
using ll = long long;

int dp[1 << 18][18];

int main() {
    int n;
    cin >> n;

    int a[2][18];
    for (int h = 0; h < 2; h++) {
        for (int i = 0; i < n; i++) {
            cin >> a[h][i];
        }
    }

    for (int k = 0; k < 1 << n; k++) {
        int x = 0;
        for (int i = 0; i < n; i++) {
            x += k >> i & 1;
        }
        if (x <= 1) continue;

        int c = 0;
        for (int i = n - 1; i >= 0; i--) {
            int t = k & 1 << i;
            if (t) {
                int s = 1 << 30;
                int l = k ^ t;
                for (int j = 0; j < n; j++) {
                    if (l & 1 << j) {
                        if (a[(x ^ j) & 1][j] <= a[(x ^ 1 ^ i) & 1][i]) s = min(s, dp[l][j] + c);
                    }
                }
                dp[k][i] = s;
                c++;
            }
        }
    }

    int r = 1 << 30;
    for (int i = 0; i < n; i++) {
        r = min(r, dp[(1 << n) - 1][i]);
    }

    if (r == 1 << 30) r = -1;
    cout << r << endl;

    return 0;
}