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(a) for a in input().split()]
B = [int(a) for a in input().split()]

def chk(L):
    NN = 5
    BIT=[0] * (2**NN+1)
    def addbit(i):
        while i <= 2**NN:
            BIT[i] += 1
            i += i & (-i)
 
    def getsum(i):
        ret = 0
        while i != 0:
            ret += BIT[i]
            i -= i&(-i)
        return ret
    s = 0
    for l in L[::-1]:
        s += getsum(l+1)
        addbit(l+1)
    return s
ans = 1 << 100
for m in range(1<<N):
    if bin(m).count("1") & 1: continue
    C = [[], []]
    for i in range(N):
        p = m >> i & 1
        q = p + i & 1
        C[q].append(((A[i] if p == 0 else B[i]) << 5) + i)
    if len(C[1]) != N // 2: continue
    C[0].sort()
    C[1].sort()
    CC = [C[i&1][i//2] for i in range(N)]
    for i in range(1, N):
        if CC[i-1] >> 5 > CC[i] >> 5: break
    else:
        CCC = [c % 32 for c in CC]
        ans = min(ans, chk(CCC))
print(ans if ans < 1 << 100 else -1)