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;

string ma(int mask, int n) {
	string ret;
	for(int i = 0; i < n; ++i) {
		if(mask >> i & 1) ret += '1';
		else ret += '0';
	}
	return ret;
}

int dp[1 << 18][55];
int main() {
	const int INF = 1e9;
	for(int i = 0; i < (1 << 18); ++i) {
		for(int j = 0; j < 55; ++j) {
			dp[i][j] = INF;
		}
	}
	dp[0][0] = 0;
	int n; cin >> n;
	int a[n][2];
	for(int i = 0; i < n; ++i) cin >> a[i][0];
	for(int i = 0; i < n; ++i) cin >> a[i][1];
	for(int mask = 0; mask < (1 << n); ++mask) {
		int cnt = __builtin_popcount(mask);
		int r = 0;
		for(int i = n - 1; i >= 0; --i) {
			if(mask >> i & 1) {
				++r;
				continue;
			}
			int nxt = a[i][abs(cnt - i) % 2];
			for(int j = 0; j <= nxt; ++j) {
				dp[mask | (1 << i)][nxt] = min(dp[mask | (1 << i)][nxt], dp[mask][j] + abs(i + r - cnt));
			}
		}
	}
	int ans = INF;
	for(int i = 0; i < 55; ++i) {
		ans = min(ans, dp[(1 << n) - 1][i]);
	}
	// for(int mask = 0; mask < (1 << n); ++mask) {
	// 	for(int i = 0; i < 5; ++i) {
	// 		cout << "(" << ma(mask, n);
	// 		printf(", %d): %d\n", i, dp[mask][i]);
	// 	}
	// }
	cout << (ans == INF ? -1 : ans) << '\n';
	return 0;
}