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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 233
const int inf=0x3f3f3f3f;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N],b[N],dp[1<<18][52],ans;
inline int Abs(int x){
	return x>=0?x:-x;
}
int main(){
	n=read();
	ans=inf;
	for(int i=1;i<=n;++i){
		a[i]=read();
	}
	for(int i=1;i<=n;++i){
		b[i]=read();
	}
	memset(dp,0x3f,sizeof(dp));
	dp[0][1]=0;
	for(int i=1;i<(1<<n);++i){
		int t=__builtin_popcount(i);
		for(int j=1;j<=n;++j){
			if((i>>j-1)&1){
				int p=(1<<j-1),num=Abs(t-j)&1?b[j]:a[j],re=0;
				for(int z=j+1;z<=n;++z){
					if((i>>z-1)&1)++re;
				}
				for(int d=0;d<=num;++d){
					dp[i][num]=min(dp[i][num],dp[i^p][d]+re);
				}
			}
		}
	}
	for(int i=1;i<=50;++i){
		ans=min(ans,dp[(1<<n)-1][i]);
	}
	printf("%d\n",ans==inf?-1:ans);
	return 0;
}