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 argc, char *argv[])
{
    int N;
    cin >> N;
    vector<int> A(N);
    vector<int> B(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    for (int i = 0; i < N; i++) {
        cin >> B[i];
    }
    vector<vector<int>> dp(N, vector<int>(1 << N, 1e9));
    for (int i = 0; i < N; i++) {
        dp[i][1 << i] = 0;
    }
    for (int mask = 0; mask < 1 << N; mask++) {
        int p = __builtin_popcount(mask);
        for (int i = 0; i < N; i++) {
            int x = i % 2 == (p - 1) % 2 ? A[i] : B[i];
            for (int j = 0; j < N; j++) {
                if (mask & 1 << j) continue;
                int y = j % 2 == p % 2 ? A[j] : B[j];
                if (x > y) continue;
                dp[j][mask | 1 << j] = min(dp[j][mask | 1 << j], dp[i][mask] + __builtin_popcount(mask >> j));
            }
        }
    }
    int res = 1e9;
    for (int i = 0; i < N; i++) {
        res = min(res, dp[i][(1 << N) - 1]);
    }
    if (res > 1e8) res = -1;
    cout << res << '\n';
    return 0;
}