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;
const int INF=1e9;
int ans=INF;
int n;
int a[20][2],p[20],t[20],b[20][2];
void jud(int s){
	for(int i=0;i<n;++i){
		if(s&(1<<i)) p[i]=1;
		else p[i]=0;
		t[i]=a[i][p[i]];
		b[i][0]=a[i][0];
		b[i][1]=a[i][1];
	}
	sort(t,t+n);
	bool flag=0;
	int val=0;
	for(int i=0;i<n;++i){
		int j=i;
		for(;j<n;++j)
			if(b[j][p[j]]==t[i]&&((j-i)&1)==p[j]) break;
		if(j>=n) {
			flag=1;
			break;
		}
		for(;j>i;--j){
			swap(b[j][0],b[j-1][1]);
			swap(b[j][1],b[j-1][0]);
			swap(p[j],p[j-1]);
			p[j]^=1;p[j-1]^=1;
			++val;
		}
	}
	if(flag) return;
	ans=min(ans,val);
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;++i) scanf("%d",&a[i][0]);
	for(int i=0;i<n;++i) scanf("%d",&a[i][1]);
	for(int s=0;s<(1<<n);++s) jud(s);
	printf("%d\n",ans==INF?-1:ans);
	return 0;
}