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 <vector>
#include <cstdlib>
#include <algorithm>
using namespace std;

int main(){
	const int INF = 1010101010;
	
	int n;
	cin >> n;
	vector<int> as(n), bs(n);
	for(int &x : as){ cin >> x; }
	for(int &x : bs){ cin >> x; }

	int p2n = 1 << n;
	vector<vector<int>> dp(p2n, vector<int>(51, INF));
	dp[0][0] = 0;
	for(int S = 0; S < p2n; ++S){
		for(int j = 1; j <= 50; ++j){
			dp[S][j] = min(dp[S][j], dp[S][j - 1]);
		}
		int ord = __builtin_popcount(S);
		for(int j = 0; j < n; ++j){
			if(S >> j & 1){ continue; }
			int val = abs(ord - j) & 1 ? bs[j] : as[j];
			int cost = __builtin_popcount(S >> j);
			int &r = dp[S | 1 << j][val];
			r = min(r, dp[S][val] + cost);
		}
	}
	int ans = *min_element(dp.back().begin(), dp.back().end());
	if(ans >= INF){ ans = -1; }
	cout << ans << '\n';
}