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;
using ll = long long;
using pii = pair<int,int>;

const int MAXN = 262150;
const ll INF = 100000000ll;

ll DP[MAXN][55], n, in[18][2];
bool vis[MAXN][55];

ll dp(int st, int val) {
	if (st == ((1 << n) - 1)) return 0;
	if (vis[st][val]) return DP[st][val];

	vis[st][val] = true;
	DP[st][val] = INF;

	vector<pii> G;
	int tmp = 0;

	for (int i = n-1; i >= 0; --i) {
		if (st & (1 << i)) tmp ^= 1;
		else G.push_back(make_pair(i, tmp));
	}

	reverse(G.begin(), G.end());

	int len = G.size(), id;
	for (int i = 0; i < len; ++i) {
		tmp = (i + G[i].second) & 1;
		id = G[i].first;
		if (in[id][tmp] >= val) {
			DP[st][val] = min(dp(st + (1 << id),in[id][tmp]) + i, DP[st][val]);
		}
	}
	return DP[st][val];
}

int main(){
	ios::sync_with_stdio(0),cin.tie(0);
	while (cin >> n) {
		for (int i = 0; i < 2; ++i) {
			for (int j = 0; j < n; ++j) {
				cin >> in[j][i];
			}
		}
		memset(vis, 0, sizeof(vis));

		if (dp(0,0) >= INF) cout << "-1\n";
		else cout << dp(0,0) << '\n';

	}
	return 0;
}