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 <iostream>
#include <string.h>
using namespace std;

const int N = 18;
int n, a[2][N], fl[N];
int f[1 << N][N][2], Ans = 10000;

void upd(int &a, int b) {
    if(a > b) a = b;
}

int main() {
	iostream::sync_with_stdio(false);
	cin >> n;
	for(int i = 0; i < n; ++ i) cin >> a[0][i];
	for(int i = 0; i < n; ++ i) cin >> a[1][i];

	memset(f, 69, sizeof f);
	for(int i = 0; i < n; ++ i) f[1 << i][i][i & 1] = i;

	for(int mask = 1; mask < (1 << n); ++ mask) {
        memset(fl, 0, sizeof fl);
        for(int i = 0; i < n; ++ i) if(mask & (1 << i))
            for(int j = 0; j < i; ++ j) fl[j] ^= 1;

		for(int i = 0; i < n; ++ i) for(int d = 0; d < 2; ++ d) if((mask & (1 << i)) && (f[mask][i][d] < 10000)) {
			if(mask == (1 << n) - 1) {
                upd(Ans, f[mask][i][d]);
                continue;
			}

            int cnt = -1;
            for(int nxt = 0; nxt < n; ++ nxt) if((mask & (1 << nxt)) == 0) {
                ++ cnt;
                int sign = fl[nxt] ^ (cnt & 1);
                if(a[d][i] <= a[sign][nxt]) upd(f[mask ^ (1 << nxt)][nxt][sign], f[mask][i][d] + cnt);
            }
		}
	}
	cout << ((Ans < 10000) ? Ans : -1) << endl;
}