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 main() {
    int n;
    cin >> n;
    vector<vector<int>> a(n, vector<int>(2));
    for (int i = 0; i < n; ++i)
        cin >> a[i][0];
    for (int i = 0; i < n; ++i)
        cin >> a[i][1];
    int c = ~0u >> 1;
    for (int m = 0; m < (1 << n); ++m) {
        vector<int> b(n);
        iota(begin(b), end(b), 0);
        int t = 0;
        auto d = [&](int i) { return a[b[i]][m >> b[i] & 1]; };
        for (int i = 0; i < n; ++i) {
            int j = i;
            for (int k = i; k < n; ++k)
                if (d(k) < d(j) ||
                    (d(k) == d(j) && ((b[j] ^ i ^ m >> b[j]) & 1)))
                    j = k;
            if ((b[j] ^ i ^ m >> b[j]) & 1) {
                t = ~0u >> 1;
                break;
            }
            int e = b[j];
            for (int k = j; k != i; --k)
                ++t, b[k] = b[k - 1];
            b[i] = e;
        }
        c = min(c, t);
    }
    cout << (c == ~0u >> 1 ? -1 : c) << endl;
    return 0;
}