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

int main(){
  int n;
  cin >> n;
  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>(51,INF));
  dp[0][0] = 0;
  rep(i,1<<n) {
    int c = 0;
    rep(j,n) {
      if(i & (1<<j)) c++;
    }
    int p = c;
    vector<P> pos;
    rep(j,n) {
       if(!(i & (1<<j))) pos.push_back(P(j,p++));
    }
    rep(l,51) {
      for(auto k : pos) {
        int d;
        if(abs(k.first-c) % 2) d = b[k.first];
        else d = a[k.first];
        if(d >= l) dp[i|(1<<k.first)][d] = min(dp[i|(1<<k.first)][d],dp[i][l]+abs(k.second-c));
      }
    }    
  }
  int ans = INF;
  rep(i,51) ans = min(ans,dp[(1<<n)-1][i]);
  cout << (ans == INF? -1 : ans) << endl;
  return 0;
}