Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds;   
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define all(a) a.begin(),a.end()
#define sz(a) (ll)(a.size())
const int M = 18;
int dp[M][(1<<M)],a[M],b[M],val[M][(1<<M)];
int n;
int getVal(int prv, int mask)
{
	int t = __builtin_popcount(mask);
	t--;
	t = abs(t - prv);
	if(t%2)
		return b[prv];
	else
		return a[prv];
}
int solve(int x,int mask)
{
	if(mask+1 == (1<<n))
		return 0;
	if(dp[x][mask] != -1)
		return dp[x][mask];
	dp[x][mask] = 1e9;
	int z = getVal(x,mask);
	for(int j=0;j<n;j++)
	{
		if((mask&(1<<j))==0)
		{
			int newMask = (mask|(1<<j));
			int y = getVal(j,newMask);
			if(y>=z)
				dp[x][mask] = min(dp[x][mask], val[j][mask]+solve(j,newMask));
		}
	}
	return dp[x][mask];
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=0;i<n;i++)
		cin>>b[i];
	memset(dp,-1,sizeof(dp));
	int ans = 1e9;
	for(int i=0;i<n;i++)
	{
		for(int mask = 0;mask<(1<<n);mask++)
		{
			if((mask&(1<<i)))
				continue;
			int z = 0;
			for(int j=0;j<n;j++)
			{
				if(j<i)
				{
					if((mask&(1<<j))==0)
						z++;
				}
				if(j>i)
				{
					if((mask&(1<<j)))
						z++;
				}
			}
			val[i][mask] = z;
		}
	}
	for(int i=0;i<n;i++)
	{
		ans = min(ans,i+solve(i,(1<<i)));
	}
	if(ans!=1e9)
		ans/=2;
	if(ans==1e9)
		ans = -1;
	cout<<ans<<"\n";
	return 0;
}