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 i64 = int64_t;

constexpr int INF = 1e9;

int main(){
    int n;
    cin >> n;
    vector<vector<int>> v(2, vector<int>());
    for(int i=0;i<2;++i){
        for(int j=0;j<n;++j){
            int tmp;
            cin >> tmp;
            v[i].push_back(tmp);
        }
    }

    int ans = INF;
    for(int i=0;i<(1<<n);++i){
        vector<int> c;
        for(int j=0;j<n;++j){
            c.push_back(v[(i>>j)&1][j]);
        }
        sort(c.begin(), c.end());

        vector<int> indices;
        deque<bool> used(n);
        for(int j=0;j<n;++j){
            for(int k=0;k<n;++k){
                if(used[k])continue;
                int d = (i>>k)&1;
                if(abs(j-k)%2 == d && c[j] == v[d][k]){
                    indices.push_back(k);
                    used[k] = true;
                    break;
                }
            }
        }

        if(indices.size() != n)continue;

        int inversion = 0;
        for(int j=0;j<n;++j){
            for(int k=j+1;k<n;++k){
                if(indices[j] > indices[k])inversion++;
            }
        }

        ans = min(ans, inversion);
    }

    cout << (ans != INF ? ans : -1) << endl;

    return 0;
}