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 <bits/stdc++.h>
using namespace std;

int a[22][2];
int main() {
    int n, m = 1e9;
    cin >> n;
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < n; j++)
            cin >> a[j][i];
    for(int i = 0; i < 1 << n; i++) {
        int x[22], d[22], e[22] = {}, b = 0;
        for(int j = 0; j < n; j++)
            x[j] = a[j][!!(i & 1 << j)];
        sort(x, x + n);
        for(int j = 0; j < n; j++) {
            for(int k = 0; k < n; k++) {
                if(e[k])
                    continue;
                if(x[j] == a[k][!!(i & 1 << k)] &&
                   (j ^ k ^ !(i & 1 << k)) & 1) {
                    d[j] = k, e[k]++;
                    goto A;
                }
            }
            goto F;
        A:;
        }
        for(int k = 0; k < n; k++)
            for(int l = k + 1; l < n; l++)
                if(d[k] > d[l])
                    b++;
        m = min(m, b);
    F:;
    }
    cout << (m < 1e9 ? m : -1) << endl;
    return 0;
}