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;
long long n, a[20], b[20], ans=999999999LL, sol;
vector<int> x; int v[20];
int main (){
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n; 
for (int i=0; i<n; i++) cin >> a[i]; 
for (int i=0; i<n; i++) cin >> b[i];

for (int i=0; i<(1<<n); i++){
	if (__builtin_popcount(i)&1) continue;
	x.clear(); for (int j=0; j<n; j++) v[j]=0;
	for (int j=0; j<n; j++) 
		if ((1<<j) & i) x.push_back(b[j]);
		else x.push_back(a[j]);
	sort(x.begin(), x.end());
	bool ok=1; sol=0;
	for (int j=0; j<n and ok; j++){
		bool kk=0;
		if ((1<<j)&i){ 
			for (int l=(j+1)%2; l<n; l+=2){
				if (x[l]==b[j] and !v[l]){
				//	cout << l << endl;
					v[l]=j+1; kk=1; break;
				}
			}
		}
		else {
			for (int l=j%2; l<n; l+=2){
				if (x[l]==a[j] and !v[l]){
				//	cout << l << endl;
					v[l]=j+1; kk=1; break;
				}
			}
		}
//	cout << i << " "; for (int i=0; i<n; i++) cout << x[i] << " "; cout << "  " << kk << "   " << sol<< endl;
		ok=ok and kk; 
	}
	if (ok){
		for (int j=0; j<n; j++) 
			for (int l=0; l<j; l++) 
			if (v[j]<v[l]) sol++;
		ans=min(ans, sol);
	}
}
if (ans==999999999LL) cout << "-1";
else cout << ans;
return 0;
}