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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int main() {
	int n = ri();
	int a[n];
	for (auto &i : a) i = ri();
	int b[n];
	for (auto &i : b) i = ri();
	
	for (int i = 1; i < n; i += 2) std::swap(a[i], b[i]);
	
	int dp[1 << n][n];
	for (auto &i : dp) for (auto &j : i) j = 1000000000;
	for (int i = 0; i < n; i++) dp[1 << i][i] = 0;
	for (int i = 0; i < 1 << n; i++) {
		for (int j = 0; j < n; j++) {
			if (dp[i][j] == 1000000000) continue;
			for (int k = 0; k < n; k++) {
				if (i >> k & 1) continue;
				if (__builtin_popcount(i) & 1) { // 2
					if (b[k] < a[j]) continue;
				} else {
					if (a[k] < b[j]) continue;
				}
				int swapped = 0;
				for (int l = k + 1; l < n; l++) if (i >> l & 1) swapped++;
				dp[i | 1 << k][k] = std::min(dp[i | 1 << k][k], dp[i][j] + swapped);
			}
		}
	}
	int res = *std::min_element(dp[(1 << n) - 1], dp[(1 << n) - 1] + n);
	if (res == 1000000000) puts("-1");
	else std::cout << res << std::endl;
	return 0;
}