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
n = int(input())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]

ans = 10 ** 18
for i in range(1 << n):
    c = []
    for j in range(n):
        if i >> j & 1 == 0:
            c.append(a[j])
        else:
            c.append(b[j])
    d = sorted(c)

    ind = {}
    for j, k in enumerate(d):
        if (j % 2, k) not in ind:
            ind[(j % 2, k)] = j

    e = []
    ok = True
    for j, k in enumerate(c):
        j = ((i >> j & 1) + j) % 2
        if (j, k) not in ind:
            ok = False
            break
        l = ind[(j, k)]
        if l >= n or d[l] != k:
            ok = False
            break
        e.append(l)
        ind[(j, k)] += 2

    if ok:
        count = 0
        for k in range(n):
            for j in range(n - 1 - k):
                if e[j] > e[j + 1]:
                    count += 1
                    e[j], e[j + 1] = e[j + 1], e[j]
        ans = min(ans, count)
if ans == 10 ** 18:
    print(-1)
else:
    print(ans)
