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<iostream>
#include<cstdio>
#include<bitset>
using namespace std;
const int N=18;

int a[N][2];
int oe[1<<N];
bitset<N*N/2>f[1<<N][N];

int main()
{
	int n,i,j,k,cnt;
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i][1]);
	for(i=0;i<n;i++)
		scanf("%d",&a[i][0]);
	for(i=1;i<(1<<n);i++)
		oe[i]=oe[i^(i&-i)]^1;
	for(i=1;i<(1<<n);i++){
		if(i==(i&-i)){
			for(j=0;j<n;j++)
				if(i&(1<<j))f[i][j][0]=1;
			continue;
		}
		for(j=0;j<n;j++)
			if(i&(1<<j)){
				cnt=0;
				for(k=n-1;k>=0;k--)
					if(i&(1<<k)){
						if(k!=j&&a[k][oe[i]^(k&1)]>=a[j][oe[i]^(j&1)^1])
							f[i][k]|=f[i^(1<<k)][j]<<cnt;
						++cnt;
					}
			}
	}
	for(i=0;i<N*N/2;i++)
		for(j=0;j<n;j++)
			if(f[(1<<n)-1][j][i]){
				printf("%d",i);
				return 0;
			}
	printf("-1");
	return 0;
}