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;
int a[109],B[109];
int dp[300009][19];
int main(){
    int n;
    cin>>n;
    if(n==1) {cout<<0; return 0;}
    for(int i =0; i < n; i++){
        cin >> a[i] ;
        dp[(1<<i)][i] = 0;
    }
    for(int i =0 ; i < n; i++)
        cin>>B[i];
    int ans = 1e9;
    for(int b = 0; b < (1<<n); b++){
        int t = 0;
          for(int j = 0; j < n; j++)
            if((1<<j)&b)
        t++;
        if(t <= 1) continue;
        for(int j = 0; j < n; j++){
            if((1<<j)&b){
                dp[b][j] = 1e9;
                int M = 1e9, inv = 0;
                for(int k = 0;k  < n; k++){
                    if(!((1<<k) &b)) continue;
                    if(k == j) continue;
                    if(j < k) inv++;
                    int x=a[j],y=a[k];
                    if(t%2 == j%2) x=B[j];
                    if(t%2 != k%2) y=B[k];
                    if(y > x) continue;
                    M= min(M, dp[b^(1<<j)][k]);

                }
                dp[b][j] = M + inv;
            }
            if(b == (1<<n)-1) ans = min(ans, dp[b][j]);
        }

    }
    if(ans == 1e9)ans=-1;
    cout<<ans<<endl;

}