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;
int N;
vector<int> A, B;
int memo[1<<18][51];
bool visited[1<<18][51];
const int INF = 1<<29;
vector<int> aa;
int f(int p, int n, int m) {
	if (p == N) {
		return 0;
	}
	if (visited[n][m]) return memo[n][m];
	visited[n][m] = true;
	int r = INF;
	for (int i = 0; i < N; ++ i) if (((n>>i)&1)==0) {
		int qq = abs(p-i) % 2;
		int x = (qq == 0) ? A[i] : B[i];
		if (x < m) continue;
		int k = 0;
		for (int j = i+1; j < N; ++ j) if ((n>>j)&1) ++ k;
		aa.push_back(i);
		r = min(r, f(p+1, n | (1<<i), x) + k);
		aa.pop_back();
	}
	return memo[n][m] = r;
}
int main() {
	cin >> N;
	A.resize(N);
	B.resize(N);
	for (auto& x : A) cin >> x;
	for (auto& x : B) cin >> x;
	int r = f(0, 0, 0);
	if (r >= INF) cout << -1 << endl;
	else cout << r << endl;
}