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 inf = 1e9;
int n;
int dp[1 << 18][55];
int A[55][2];
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> A[i][0];
    for (int i = 1; i <= n; i++) cin >> A[i][1];
    for (int i = 1; i < (1 << n); i++) {
        int x = __builtin_popcount(i);
        for (int j = 1; j <= 50; j++) {
            dp[i][j] = inf;
        }
        for (int j = 1; j <= n; j++) {
            if (i & (1 << j - 1)) {
                int v = A[j][x - j & 1];
                dp[i][v] = min(dp[i][v], dp[i ^ (1 << j - 1)][v] + __builtin_popcount(i ^ (i & (1 << j) - 1)));
            }
        }
        for (int j = 2; j <= 50; j++) {
            dp[i][j] = min(dp[i][j], dp[i][j - 1]);
        }
    }
    cout << (dp[(1 << n) - 1][50] < inf ? dp[(1 << n) - 1][50] : -1);
}