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 N=18,inf=0x1f1f1f1f;
int a[N],b[N],f[1<<N][55];
int main(){
	int n,s;scanf("%d",&n),s=(1<<n);
	for(int i=0;i<n;i++)scanf("%d",a+i);
	for(int i=0;i<n;i++)scanf("%d",b+i);
	memset(f,0x1f,sizeof(f));
	for(int i=0;i<=50;i++)f[0][i]=0;
	for(int i=1;i<s;i++){
		int w=0,w1=0;
		for(int j=0;j<n;j++)if(i&(1<<j))w++;
		for(int j=0;j<n;j++){
			if(!(i&(1<<j)))continue;
			if((w-j)&1)f[i][a[j]]=min(f[i][a[j]],f[i^(1<<j)][a[j]]+j-w1);
			else f[i][b[j]]=min(f[i][b[j]],f[i^(1<<j)][b[j]]+j-w1);
			w1++; 
		}
		for(int j=1;j<=50;j++)f[i][j]=min(f[i][j],f[i][j-1]);
	}
	printf("%d",f[s-1][50]==inf?-1:f[s-1][50]);
	return 0;
}