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 main(){
    int n;
    cin >> n;

    vector<int> a(n), 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((1<<n), vector<int>(51, INF));
    dp[0][0] = 0;
    for(int bit=0; bit<(1<<n); bit++){

        for(int xi=0; xi<=50; xi++){
            if(dp[bit][xi] == INF) continue;

            int pos = __builtin_popcount(bit);
            for(int j=0; j<n; j++){
                if(bit >> j & 1) continue;

                int xj = a[j];
                if(abs(j - __builtin_popcount(bit)) % 2 == 1) xj = b[j];

                if(xj >= xi) dp[bit|(1<<j)][xj] = min(dp[bit|(1<<j)][xj], dp[bit][xi] + (pos - __builtin_popcount(bit)));
                pos += 1;
            }
        }
    }

    /*
    for(int bit=0; bit<(1<<n); bit++){
        for(int xi=0; xi<=5; xi++){
            cout << setw(11) << dp[bit][xi];
        }
        cout << endl;
    }
    */

    int ans = INF;
    for(int xi=0; xi<=50; xi++){
        ans = min(ans, dp[(1<<n)-1][xi]);
    }
    if(ans != INF) cout << ans << endl;
    if(ans == INF) cout << -1  << endl;
    return 0;
}