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>
const int N=20,S=27e4;
int f[N][S][N],n,i,j,k,l,a[2][N],cnt[S],val[N][N],ans;
inline void upi(int&a,const int&b){a>b?a=b:0;}
int main(){
	scanf("%d",&n);
	for(i=1;i<1<<n;++i)cnt[i]=cnt[i>>1]+(i&1);
	for(j=0;j<2;++j)for(i=0;i<n;++i)scanf("%d",a[j]+i);
	memset(f,1,sizeof f);for(i=0;i<n;++i)f[0][1<<i][i]=0;ans=1<<25;
	for(i=0;i<n;++i)for(j=0;j<n;++j)val[i][j]=a[(i^j)&1][j];
	for(i=0;i+1<n;++i)for(j=0;j<1<<n;++j)if(cnt[j]==i+1)for(k=0;k<n;++k)if(j>>k&1)
		for(l=0;l<n;++l)if(!(j>>l&1) && val[i][k]<=val[i+1][l])
			upi(f[i+1][j|(1<<l)][l],f[i][j][k]+cnt[j&~((1<<l)-1)]);
	for(i=0;i<n;++i)upi(ans,f[n-1][(1<<n)-1][i]);
	if(ans>114514)ans=-1;
	printf("%d\n",ans);
}