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;
const int M = 51;
const int MOD = 1e9+7;
typedef pair<int, int> pii;

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

inline void update(int &x, const int &y)
{
    if (x == -1 || y < x) x = y;
}

inline void solve()
{
    cin >> n;
    for (int j = 0; j < 2; ++j)
        for (int i = 0; i < n; ++i)
            cin >> a[i][j];

    memset(dp, -1, sizeof dp);
    dp[0][1] = 0;
    for (int i = 1; i < 1<<n; ++i) {
        int t = __builtin_popcount(i)-1;
        for (int p = 0; p < n; ++p) if ((i>>p)&1) {
            int j = a[p][abs(t-p)&1], num = 0;
            for (int k = p+1; k < n; ++k) num += (i>>k)&1;
            for (int k = 1; k <= j; ++k) if (dp[i^(1<<p)][k] != -1) {
                update(dp[i][j], dp[i^(1<<p)][k]+num);
            }
        }
    }
    int res = -1;
    for (int j = 0, i = (1<<n)-1; j < M; ++j) if (dp[i][j] != -1) {
        update(res, dp[i][j]);
    }
    cout << res << endl;
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int T = 1;
    while (T--) {
        solve();
    }
    return 0;
}