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;
typedef long long ll;
typedef pair<int,int> PII;
const int maxn=100010;
#define MP make_pair
#define PB push_back
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline ll read(){
	char ch=getchar();ll x=0,f=0;
	while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
	return f?-x:x;
}
int n,a[2][18],f[18][262144],cnt[18][262144],sz[262144],ans=1e9;
int main(){
	n=read();
	FOR(i,0,n-1) a[0][i]=read();
	FOR(i,0,n-1) a[1][i]=read();
	FOR(i,0,(1<<n)-1) sz[i]=sz[i>>1]+(i&1);
	FOR(i,0,n-1) FOR(j,0,(1<<n)-1) FOR(k,0,n-1) if(((j>>k)&1) && i>k) cnt[i][j]++;
	MEM(f,0x3f);
	FOR(i,0,n-1) f[i][1<<i]=cnt[i][(1<<n)-1];
	FOR(j,1,(1<<n)-1) FOR(i,0,n-1) if((j>>i)&1){
		FOR(k,0,n-1) if(!((j>>k)&1) && a[abs(i-(sz[j]-1))%2][i]<=a[abs(k-sz[j])%2][k]){
			f[k][j|(1<<k)]=min(f[k][j|(1<<k)],f[i][j]+cnt[k][((1<<n)-1)^j]);
		}
	}
	FOR(i,0,n-1) ans=min(ans,f[i][(1<<n)-1]);
	printf("%d\n",ans==1e9?-1:ans);
}