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;
using ll = long long;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
//#include "library/debug.cpp"

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;
    vector<vector<int>> a(n, vector<int>(2));
    rep(i, 0, n) cin >> a[i][0];
    rep(i, 0, n) cin >> a[i][1];
    for (int i = 1; i < n; i += 2) swap(a[i][0], a[i][1]);
    vector<vector<int>> dp(1 << n, vector<int>(n, 1e9));
    rep(i, 0, n) {
        dp[1 << i][i] = 0;
    }

    for (int x = 1; x < (1 << n) - 1; x++) {
        int cnt = (__builtin_popcount(x) + 1) % 2;
        rep(i, 0, n) if ((x >> i) & 1) {
            if (dp[x][i] > 1e7) continue;
            int mn = a[i][cnt];
            rep(j, 0, n) if (!((x >> j) & 1)) {
                if (mn <= a[j][cnt ^ 1]) {
                    int res = 0;
                    rep(k, j + 1, n) if ((x >> k) & 1) res++;
                    dp[x + (1 << j)][j] =
                        min(dp[x + (1 << j)][j], dp[x][i] + res);
                }
            }
        }
    }
    int ans = 1e9;
    rep(i, 0, n) ans = min(ans, dp[(1 << n) - 1][i]);
    cout << (ans > 1e7 ? -1 : ans) << endl;
    return 0;
}