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;
int a[18][2];
int c[19];
int dp[19][1 << 18][18];
int main() {
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) scanf("%d", &a[i][0]);
	for(int i(0); i < n; i++) {
		scanf("%d", &a[i][1]);
		if(i % 2 == 1) swap(a[i][0], a[i][1]);
	}
	memset(dp, 0x7f, sizeof(dp));
	for(int i(0); i < n; i++) {
		dp[0][1 << i][i] = 0;
	}
	for(int i(0); i < n - 1; i++) {
		for(int j(0); j < (1 << n); j++) {
			if(__builtin_popcount(j) != i + 1) continue;
			c[n] = 0;
			for(int k = n - 1; k >= 0; k--) {
				c[k] = c[k + 1] + (j >> k & 1);
			}
			for(int k = 0; k < n; k++) {
				if(!(j >> k & 1)) continue;
				int lst = i % 2 == 0 ? a[k][0] : a[k][1];
				//cout << i << ' ' << j << ' ' << k << ' ' << lst << endl;
				for(int l = 0; l < n; l++) {
					if(j >> l & 1) continue;
					int cur = i % 2 == 0 ? a[l][1] : a[l][0];
					if(cur >= lst) {
						//cout << "->" << l << endl;
						dp[i + 1][j + (1 << l)][l] = min(dp[i + 1][j + (1 << l)][l], dp[i][j][k] + c[l]);
					}
				}
			}
		}
	}
	int ans = 0x7f7f7f7f;
	for(int i(0); i < n; i++) ans = min(ans, dp[n - 1][(1 << n) - 1][i]);
	if(ans == 0x7f7f7f7f) ans = -1;
	cout << ans << endl;
}