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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template<class T,class U> using P = pair<T,U>;
template<class T> using vec = vector<T>;
template<class T> using vvec = vector<vec<T>>;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vec<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];
    int inf = 1e9;
    vvec<int> dp(1<<N,vec<int>(51,inf));
    dp[0][0] = 0;
    for(int S=0;S<(1<<N);S++){
        int cost = 0;
        for(int j=0;j<N;j++){
            if(!(S&(1<<j))){
                int n = __builtin_popcount(S);
                int ne = (abs(n-j)%2? B[j]:A[j]);
                int T = S^(1<<j);
                for(int k=0;k<=50;k++) if(k<=ne && dp[S][k]!=inf){
                    dp[T][ne] = min(dp[T][ne],dp[S][k]+cost);
                }
                cost++;
            }
        }
    }
    int ans = inf;
    for(int k=0;k<=50;k++) ans = min(ans,dp[(1<<N)-1][k]);
    cout << (ans==inf? -1:ans) << endl;
}