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 <algorithm>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<int> A(N);
    for (int &a: A)
        cin>>a;
    vector<int> B(N);
    for (int &b: B)
        cin>>b;

    int ans = 0x7fffffff;

    vector<int> T(N), X(N);
    for (int b=0; b<1<<N; b++)
    {
        for (int i=0; i<N; i++)
            T[i] = b>>i&1 ? B[i] : A[i];
        sort(T.begin(), T.end());

        for (int i=0; i<N; i++)
            X[i] = i;
        int a = 0;
        bool ok = true;
        for (int i=0; i<N; i++)
        {
            int p;
            for (p=i; p<N; p++)
                if ((b>>X[p]&1 ? B[X[p]] : A[X[p]])==T[i] &&
                    ((X[p]-i)&1)==(b>>X[p]&1))
                    break;
            if (p>=N)
            {
                ok = false;
                break;
            }

            for (int j=p-1; j>=i; j--)
            {
                swap(X[j], X[j+1]);
                a++;
            }
        }
        if (ok)
            ans = min(ans, a);
    }
    cout<<(ans < 0x7fffffff ? ans : -1)<<endl;
}