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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,a[20],b[20],dp[1<<18][51],cnt[1<<18],ans=0x3f3f3f3f;
int main()
{
	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]);
	for(int i=1;i<1<<n;++i)cnt[i]=cnt[i>>1]+(i&1);
	memset(dp,0x3f,sizeof(dp));dp[0][1]=0;
	for(int i=0;i<(1<<n)-1;++i)
		for(int j=1;j<=50;++j)
			if(dp[i][j]!=0x3f3f3f3f)
				for(int k=0,w,c=cnt[i];k<n;++k)
				{
					w=(k&1)==(cnt[i]&1)?a[k]:b[k];
					if(i>>k&1)--c;
					if(!(i>>k&1)&&w>=j)dp[i|(1<<k)][w]=min(dp[i|(1<<k)][w],dp[i][j]+c);
				}
//	for(int i=0;i<1<<n;++i)
//		for(int j=1;j<=3;++j)
//			printf("%d %d %d\n",i,j,dp[i][j]);
	for(int i=1;i<=50;++i)ans=min(ans,dp[(1<<n)-1][i]);
	printf("%d",ans==0x3f3f3f3f?-1:ans);
	return 0;
}