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 <string>
#include <array>
#include <vector>
#include <algorithm>

using namespace std;
static constexpr int INF = 1000000000;

int main(){
	ios_base::sync_with_stdio(false);
	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]; }
	vector<array<int, 51>> dp(1 << n);
	for(auto& a : dp){ fill(a.begin(), a.end(), INF); }
	dp[0][0] = 0;
	for(int i = 0; i < (1 << n); ++i){
		for(int j = 0; j <= 50; ++j){
			for(int k = 0, c = 0; k < n; ++k){
				if(i & (1 << k)){ continue; }
				const int flip = (c ^ __builtin_popcount(i & ~((1 << k) - 1))) & 1;
				const int v = flip ? b[k] : a[k];
				if(v >= j){ dp[i | (1 << k)][v] = min(dp[i | (1 << k)][v], dp[i][j] + c); }
				++c;
			}
		}
	}
	int answer = INF;
	for(int i = 0; i <= 50; ++i){
		answer = min(answer, dp.back()[i]);
	}
	if(answer >= INF){ answer = -1; }
	cout << answer << endl;
	return 0;
}