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;

const int N = 18, INF = 1005;

int n, a[N][2], dp[1 << N][N];

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i][1];
    }
    for (int i = 0; i < n; i++) {
        cin >> a[i][0];
    }
    for (int msk = 0; msk < (1 << n); msk++) {
        for (int i = 0; i < n; i++) {
            dp[msk][i] = INF;
        }
    }
    for (int msk = 1; msk < (1 << n); msk++) {
        int les = 0, cnt = __builtin_popcount(msk);
        for (int i = 0; i < n; i++) {
            if (msk >> i & 1) {
                if (cnt == 1) {
                    dp[msk][i] = 0;
                }
                dp[msk][i] += les;
                int cur = a[i][(cnt ^ i) & 1];
                for (int j = 0; j < n; j++) {
                    if (!(msk >> j & 1) && a[j][(cnt ^ j ^ 1) & 1] >= cur) {
                        dp[msk ^ (1 << j)][j] = min(dp[msk ^ (1 << j)][j], dp[msk][i]);
                    }
                }
            } else {
                les++;
            }
        }
    }
    int mi = *min_element(dp[(1 << n) - 1], dp[(1 << n) - 1] + n);
    cout << (mi >= INF ? -1 : mi);
}