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;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
typedef pair<int, int> P;
int a[20], b[20];

int dp[1 << 18][18];
int popcnt(int n) {
	if (!n)return 0;
	return (n & 1) + popcnt(n >> 1);
}
signed main() {
	int n; cin >> n;
	rep(i, n)cin >> a[i]; rep(i, n)cin >> b[i];
	rep(j, 1 << n)rep(i, n)dp[j][i] = 1e9;
	rep(i, n)dp[1 << i][i] = i;

	rep(i, 1 << n) {
		rep(j, n) {
			if (i&(1 << j)) {
				int now;
				if (abs(j - popcnt(i)) & 1)now = a[j];
				else now = b[j];
				rep(k, n) {
					if (!(i&(1 << k))) {//kを追加
						int po = popcnt(i);
						int nxt;
						if (abs(k - po) & 1)nxt = b[k];
						else nxt = a[k];
						if (now <= nxt) {
							int cst = k-popcnt(i&((1 << k) - 1));
							dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + cst);
						}
					}
				}
			}
		}
	}
	int res = 1e9;
	rep(i, n)res = min(res, dp[(1 << n) - 1][i]);
	if (res == 1e9)res = -1;
	cout << res << endl;
}