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

INF = N *100
ans = INF
for i in range(2**N):
    C = [((a,j,0) if (i>>j)&1 else (b,j,1)) for j,a,b in zip(range(N),A,B)]
    P = sum(p for c,j,p in C)
    if P%2: continue
    E = sum(p for c,j,p in C[::2])
    O = sum(p for c,j,p in C[1::2])
    if E!=O: continue
    C0 = []
    C1 = []
    for c,j,p in C:
        if (p+j)%2:
            C1.append((c,j,p))
        else:
            C0.append((c,j,p))
    C0.sort()
    C1.sort()
    D = []
    E = []
    for c0,c1 in zip(C0,C1):
        E.append(c0[0])
        E.append(c1[0])
        D.append(c0[1])
        D.append(c1[1])
    if N%2:
        D.append(C0[-1][1])
        E.append(C0[-1][0])
    contFlg = False
    for e,e2 in zip(E[:-1],E[1:]):
        if e>e2:
            contFlg = True
            break
    if contFlg: continue
    tmp = 0
    for i,d in enumerate(D):
        for j,d2 in enumerate(D[:i]):
            if d2 > d: tmp += 1
    ans = min(ans,tmp)
if ans == INF:
    ans = -1
print(ans)