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;
const int maxn = 1 << 20, inf = 0x3f3f3f3f;
int n;
int dp[maxn][19][2], a[20][2], pos[maxn][20];
int main() {
	cin >> n;
	for(int i = 0; i < n; ++i) {
		cin >> a[i][0];
	}
	for(int i = 0; i < n; ++i) {
		cin >> a[i][1];
	}
	memset(dp, 0x3f3f, sizeof(dp));
	for(int i = 0; i < n; ++i) {
		dp[1 << i][i][i & 1] = i;
	}
	for(int S = 0; S < 1 << n; ++S) {
		for(int i = 0; i < n; ++i) {
			pos[S][i] = i;
			for(int j = i + 1; j < n; ++j) {
				if(S >> j & 1) {
					++pos[S][i];
				}
			}
		}
	}
	for(int S = 0; S < 1 << n; ++S) {
		int c = __builtin_popcount(S);
		for(int i = 0; i < n; ++i) {
			if(S >> i & 1) {
				for(int j = 0; j < n; ++j) {
					if(!(S >> j & 1)) {
						for(int k = 0; k < 2; ++k) {
							if(a[j][abs(j - c) & 1] >= a[i][k]) {
								dp[S ^ (1 << j)][j][abs(j - c) & 1] = min(dp[S ^ (1 << j)][j][abs(j - c) & 1], dp[S][i][k] + pos[S][j] - c);
							}
						}
					}
				}
			}
		}
	}
	int ans = inf;
	for(int i = 0; i < n; ++i) {
		for(int j = 0; j < 2; ++j) {
			ans = min(ans, dp[(1 << n) - 1][i][j]);
		}
	}
	cout << (ans == inf ? -1 : ans) << endl;
	/*
		1 2 3 -> 1 -3 -2 -> 3 -1 -2
		1 2 3 -> -2 -1 3 -> -2 -3 1
		dp[S][i][0][1] : 选中了S集合 以i为结尾最少步数
	*/
	return 0;
}