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 = list(map(int, input().split()))
B = list(map(int, input().split()))

v = [[0, 0] for _ in range(N)]
s = []
ans = 500

for i in range(1<<N):
    tmp = 0
    for j in range(N):
        if ((i>>j)&1): v[j] = [B[j], j]
        else:          v[j] = [A[j], j]
    for j in range(N):
        p = j
        while (p and v[p][0] < v[p-1][0]):
            v[p], v[p-1] = v[p-1], v[p]
            tmp += 1
            p -= 1
    for j in range(N):
        if (((i>>v[j][1])+v[j][1]-j)&1):
            s.append(v[j][0]*500+j)
    if (len(s)&1):
        tmp = 500
        s.pop()
    while s:
        p = s.pop()
        q = s.pop()
        if ((p-q)&1): tmp += p-q
        else:         tmp += 500
    ans = min(ans, tmp)

if ans < 500:
    print(ans)
else:
    print(-1)
