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()))

A = [A,B]

import itertools

C = list(itertools.combinations([i for i in range(n)], n//2))
D = [[0] * 2 for i in range(n)]


def sw(G):
    a = 0
    # print("G",G)
    for i in range(len(G)):
        j = i
        while G[j] != i:
            j += 1
        for k in range(j, i, -1):
            ss = G[k]
            G[k] = G[k-1]
            G[k-1] = ss
            a += 1
            # print(G)
    # print(a)
    return a

ans = float("inf")

for i in range(len(C)):
    D = [[0] * 2 for i in range(n//2)]
    E = [[0] * 2 for i in range(n - n//2)]
    L = [0] * n
    for k in range(len(C[i])):
        D[k][0] = A[(C[i][k]+1)%2][C[i][k]]
        D[k][1] = C[i][k]
        L[C[i][k]] = 1
    l = 0
    for k in range(n):
        if L[k] == 0:
            E[l][0] = A[k%2][k]
            E[l][1] = k
            l += 1
    D.sort(key=lambda x:x[1])
    D.sort()
    E.sort(key=lambda x:x[1])
    E.sort()
    F = []
    for i in range(n):
        if i % 2 == 0:
            F.append(E[i//2])
        else:
            F.append(D[i//2])
    # print(F)
    ok = 1
    for i in range(len(F)-1):
        if F[i][0] > F[i+1][0]:
            ok = 0
    # print("ok",ok)
    if ok == 1:
        ans0 = sw([F[i][1] for i in range(n)])
        ans = min(ans, ans0)
    
if ans == float("inf"):
    ans = -1
print(ans)
    


