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 V=50;
const int MX=1e9+7;
int n,a[20],b[20],off[20][20],dp[1<<20][50];
int fun(int mask,int val)
{
    int ans=MX,d=0;
    if(mask==(1<<n)-1) return 0;
    if(dp[mask][val]!=-1) return dp[mask][val];
    int ln=__builtin_popcount(mask);
    for(int i=0;i<n;i++)
    {
        if(mask&(1<<i)) continue;
        if(off[ln][i]>=val)
        ans=min(ans,d+fun(mask|(1<<i),off[ln][i]));
        d++;
    }
    return dp[mask][val]=ans;
}
int main()
{
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i];
    for(int i=0;i<n;i++) cin>>b[i];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        off[i][j]=(i%2==j%2)?a[j]:b[j];
    }
    memset(dp,-1,sizeof(dp));
    int res=fun(0,0);
    if(res>=MX) cout<<"-1";
    else
    cout<<res;
    return 0;
}