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 F first
#define S second
#define pb push_back
using namespace std;

typedef long long ll;

const int N = 19;
const int MAXN = (1<<N);
const int MAXVAL = 1e9;

int dp[N + 1][MAXN];

int A[N + 5], B[N + 5];
int n;

int solve(int pre, int bmask) {
	if(bmask == 0) return 0;
	int &res = dp[pre][bmask];
	if(res != -1) return res;
	res = MAXVAL;
	int getID = n + 1 - __builtin_popcount(bmask);
	int pval = A[pre];
	if((getID + pre + 1) % 2) pval = B[pre];
	int pass = getID - 1;
	for(int i = 1;i <= n;i++) {
		if(bmask & (1<<i)) {
			int curval = A[i];
			if((getID + i) % 2) curval = B[i];
			if(pval > curval) continue;
			res = min(res, solve(i, bmask ^ (1<<i)) + (i + pass - getID));
		} else pass--;
	}
	return res;
}

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin>>n;
	memset(dp, -1, sizeof dp);
	for(int i = 1;i <= n;i++) cin>>A[i];
	for(int i = 1;i <= n;i++) cin>>B[i];
	int result = solve(0, (1<<(n + 1)) - 2);
	if(result >= MAXVAL) result = -1;
	cout<<result<<endl;
}