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>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)x.size())
#define FOR(i,a,b) for (int i=a;i<=b;++i)
#define FORD(i,a,b) for (int i=a;i>=b;--i)
using namespace std;
typedef long long LL;
typedef pair<int,int> pa;
typedef vector<int> vec;
void getint(int &v){
    char ch,fu=0;
    for(ch='*'; (ch<'0'||ch>'9')&&ch!='-'; ch=getchar());
    if(ch=='-') fu=1, ch=getchar();
    for(v=0; ch>='0'&&ch<='9'; ch=getchar()) v=v*10+ch-'0';
    if(fu) v=-v;
}
const int INF=1e9+10;
int n,pc[500010],f[300010][55],ans,a[500010],b[500010];
int tr(int x,int y){
	if ((abs(x-y))&1) return b[x]; else return a[x];
}
int main(){
	pc[0]=0;
	FOR(i,1,500000) pc[i]=pc[i>>1]+(i&1);
	cin>>n;
	FOR(i,0,n-1) getint(a[i]);
	FOR(i,0,n-1) getint(b[i]);
	FOR(i,0,(1<<n)-1) FOR(j,1,50) f[i][j]=INF;
	f[0][1]=0;
	FOR(i,0,(1<<n)-2)
		FOR(j,1,50)
			if (f[i][j]<INF){
				int cnt=0;
				FOR(k,0,n-1){
					if ((i>>k)&1){++cnt;continue;}
					int too=tr(k,pc[i]);
					if (too>=j) f[i|(1<<k)][too]=min(f[i|(1<<k)][too],f[i][j]+pc[i]-cnt);
				}
			}
	ans=INF;
	FOR(i,1,50) ans=min(ans,f[(1<<n)-1][i]);
	if (ans<INF) cout<<ans<<endl; else puts("-1");
	return 0;
}