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=20,M=2e6,inf=1e9;
int n,a[N],b[N],mx,f[N][M],bit[M],ans=1e9;
void cmin(int &x,int y){y<x?x=y:0;}
int main(){
	scanf("%d",&n);mx=1<<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=0;i<n;i++)if(i&1)swap(a[i],b[i]);
	for(int i=1;i<mx;i++)bit[i]=bit[i>>1]+(i&1);
	memset(f,0x3f,sizeof(f));
	for(int i=0;i<n;i++)f[i][1<<i]=0;
	for(int i=1;i<mx;i++)
		for(int j=n-1,l=0;~j;j--)
			if(!(i>>j&1)){
				int y=bit[i]&1?b[j]:a[j];
				for(int k=0;k<n;k++)if(i>>k&1){
					int x=bit[i]&1?a[k]:b[k];
					if(x<=y)cmin(f[j][i^1<<j],f[k][i]+l);
				}
			}
			else l++;
	for(int i=0;i<n;i++)cmin(ans,f[i][mx-1]);
	if(ans>n*n)puts("-1");
	else printf("%d\n",ans);
}