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 mn=1e5+10;
typedef pair<int,int>pii;
int a[mn][2];
int dp[1<<18][18];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",&a[i][0]);
    for(int i=0;i<n;i++)scanf("%d",&a[i][1]);
    for(int i=1;i<n;i+=2)swap(a[i][0],a[i][1]);
    memset(dp,0x3f,sizeof(dp));
    for(int i=0;i<n;i++)dp[1<<i][i]=i;
    for(int i=1;i<1<<n;i++){
        if(__builtin_popcount(i)==1)continue;
        for(int j=0;j<n;j++)if((i>>j)&1){
            int mov=j-__builtin_popcount(i%(1<<j));
            int ind=__builtin_popcount(i)&1^1;
            for(int k=0;k<n;k++){
                if(j==k)continue;
                if(a[k][ind^1]<=a[j][ind])dp[i][j]=min(dp[i][j],dp[i-(1<<j)][k]+mov);
            }
        }
    }
    int ans=0x3f3f3f3f;
    for(int i=0;i<n;i++)ans=min(ans,dp[(1<<n)-1][i]);
    if(ans<1000)printf("%d",ans);
    else printf("-1");
}