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 long long ll;
const int max_n=20;
const int inf=0x3f3f3f3f;
int n,a[max_n],b[max_n];
int dp[1<<18][max_n];
int get(int x,int y){
    return (x^y)&1?b[x]:a[x];
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",a+i);
    for(int i=0;i<n;i++)scanf("%d",b+i);
    for(int s=1;s<(1<<n);s++){
        if(s&(s-1)){
            int c=__builtin_popcount(s);
            for(int i=0;i<n;i++){
                if(s>>i&1){
                    dp[s][i]=inf;
                    int t=s^(1<<i);
                    int cost=0;
                    for(int j=i+1;j<n;j++)if(s>>j&1)cost++;
                    for(int j=0;j<n;j++){
                        if((t>>j&1)&&get(j,c-2)<=get(i,c-1)){
                            dp[s][i]=min(dp[s][i],dp[t][j]+cost);
                        }
                    }
                }
            }
        }
    }
    int ans=inf;
    for(int i=0;i<n;i++)ans=min(ans,dp[(1<<n)-1][i]);
    if(ans!=inf)printf("%d\n",ans);
    else printf("-1\n");
    return 0;
}