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;
typedef pair<int,int> Pii;
#define rep(i,n) for (int i=0;i<(n);++i)
const int INF=1e9;

template<class T> inline bool chmin(T &a,T b){
    if (a>b){a=b; return true;} return false;
}

const int MAX_AB=51;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin >> N;
    vector<int> A(N),B(N);
    rep(i,N) cin >> A[i];
    rep(i,N) cin >> B[i];
    vector<vector<int>> dp(1<<N,vector<int>(MAX_AB,INF));
    dp[0][0]=0;
    rep(bit,1<<N){
        int cnt=0;
        vector<Pii> can;
        rep(i,N) if (bit&1<<i) ++cnt;
        int order=cnt;
        rep(i,N) if (!(bit&1<<i)) can.emplace_back(i,order++);
        rep(j,MAX_AB){
            if (dp[bit][j]>=INF) continue;
            for (Pii p:can){
                int i=p.first,nj=(i%2==cnt%2?A[i]:B[i]);
                if (nj>=j) chmin(dp[bit|1<<i][nj],dp[bit][j]+p.second-cnt);
            }
        }
    }
    int ans=INF;
    rep(j,MAX_AB) chmin(ans,dp[(1<<N)-1][j]);
    if (ans==INF) ans=-1;
    cout << ans << endl;
}