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;
const int inf=0x3f3f3f3f,maxn=1<<18|10;
int dp[maxn][60];
int a[20],b[20];
int n,S;
void solve(){
    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);
    memset(dp,0x3f,sizeof dp);
    S=1<<n;
    dp[0][0]=0;
    for(int s=0,cnt;s<S;++s){
        cnt=__builtin_popcount(s);
        for(int j=0,t,u,p=0;j<n;++j)if(~s&(1<<j)){
            u=(abs(cnt-j)&1)?b[j]:a[j];
            for(int k=0;k<=50;++k)
                if(u>=k)dp[s|1<<j][u]=min(dp[s|1<<j][u],dp[s][k]+p);
            ++p;
        }
    }
    int ans=inf;
    for(int j=0;j<=50;++j)
        ans=min(ans,dp[S-1][j]);
    if(ans==inf)puts("-1");
    else printf("%d\n",ans);
}
int main(){
    solve();

    return 0;
}