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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000

int main(){
	
	int N;
	cin>>N;
	
	vector<int> A(N),B(N);
	
	for(int i=0;i<N;i++){
		cin>>A[i];
	}
	
	for(int i=0;i<N;i++){
		cin>>B[i];
	}
	
	int ans = Inf;
	
	for(int i=0;i<(1<<N);i++){
		vector<int> C;
		for(int j=0;j<N;j++){
			if((1<<j)&i)C.push_back(B[j]);
			else C.push_back(A[j]);
		}
		
		sort(C.begin(),C.end());
		
/*
		for(int j=0;j<C.size();j++){
			cout<<C[j]<<',';
		}
		cout<<endl;
	*/	
		vector<bool> used(N,false);
		
		vector<int> ind;
		
		for(int j=0;j<N;j++){
			for(int k=0;k<N;k++){
				if(used[k])continue;
				if((1<<k)&i){
					if((abs(j-k)%2==1)&&B[k] == C[j]){
						used[k] = true;
						ind.push_back(k);
						break;
					}
				}
				else{
					if((abs(j-k)%2==0)&&A[k] == C[j]){
						used[k] = true;
						ind.push_back(k);
						break;
					}
				}
			}
		}
		
		
		if(ind.size()!=N)continue;
		int tmp = 0;
		for(int j=0;j<N;j++){
			for(int k=0;k<j;k++){
				if(ind[k] > ind[j])tmp++;
			}
		}
		
		//if(i==0)cout<<tmp<<endl;
		
		ans = min(ans,tmp);
		
	}
	if(ans == Inf)cout<<-1<<endl;
	else cout<<ans<<endl;
	
	return 0;
}