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 LL long long
#define PII pair<int, int>
#define f first
#define s second 
using namespace std;
const LL INF = (LL) 1e9 + 5;
const LL MOD = (LL) 1e9 + 7;
const int MAXN = 25;

int n, dp[1 << 18][MAXN], num[2][MAXN], odd[1 << 18][MAXN];

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);

	cin >> n;
	for (int i = 0; i < n; i++) cin >> num[0][i];
	for (int i = 0; i < n; i++) cin >> num[1][i];
	
	for (int i = 0; i < (1 << n); i++) {
		for (int j = 0; j < n; j++) {
			int res = 0;
			for (int k = 0; k < n; k++) {
				if (k < j && (i >> k & 1)) res ^= 1;
				if (k > j && (~i >> k & 1)) res ^= 1;
			}
			odd[i][j] = res;
		}
	}
	
	for (int i = 1; i < (1 << n) - 1; i++) {
		for (int j = 0; j < n; j++) {
			dp[i][j] = INF;
			if (i >> j & 1) continue;
			int pre = num[odd[i][j]][j], mv = 0;
			for (int k = 0; k < n; k++) {
				if (~i >> k & 1) continue;
				int x = num[odd[i][k]][k];
				if (x >= pre) {
					dp[i][j] = min(dp[i][j], dp[i ^ 1 << k][k] + mv);
				}
				mv++;
			}
		}
	}
	
	int ans = INF, all = (1 << n) - 1;
	for (int i = 0; i < n; i++) {
		ans = min(ans, dp[all ^ 1 << i][i] + i);
	}
	
	cout << (ans >= INF ? -1 : ans) << '\n';

	return 0;
}