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;

struct Vec {
        int x, y, n, t;
        bool operator < (const Vec &temp) const {
                return x < temp.x;
        }
};

int n, a[19][2], ans = 2e9;

signed main () {
	ios::sync_with_stdio(0);
	cin.tie(0);

	cin >> n;
	for(int k = 0; k <= 1; ++k)
		for(int i = 1; i <= n; ++i) cin >> a[i][k];
	for(int mask = 0; mask < (1 << n); ++mask) {
		vector<Vec> v(n + 1);
		int moves = 0, step = n;
		for(int i = 1; i <= n; ++i) v[i] = {a[i][0], a[i][1], 0, ((mask >> (i - 1)) & 1)};
		while(step > 0) {
			int mx = 0, pos = 0;
			for(int i = 1; i <= step; ++i) {
				int after = ((step - i) % 2) ^ v[i].n;
				if(after != v[i].t) continue;
				int destval = (after == v[i].n ? v[i].x : v[i].y);
				if(destval >= mx) mx = destval, pos = i;
			}
			if(pos == 0) {
                moves = 2e9;
				break;
			}
			for (int i = pos; i < step; ++i, ++moves) {
				swap(v[i].x, v[i].y);
				v[i].n ^= 1;
				swap(v[i + 1].x, v[i + 1].y);
				v[i + 1].n ^= 1;
				swap(v[i], v[i + 1]);
			}
			--step;
		}
		if(is_sorted(v.begin() + 1, v.end())) ans = min(ans, moves);
	}
	cout << (ans == 2e9 ? -1 : ans);
}