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
from itertools import combinations
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))

def inversion(inds):
    bit = [0] * (N+1)
    def bit_add(x,w):
        while x <= N:
            bit[x] += w
            x += (x & -x)
    def bit_sum(x):
        ret = 0
        while x > 0:
            ret += bit[x]
            x -= (x & -x)
        return ret
    inv = 0
    for ind in reversed(inds):
        inv += bit_sum(ind + 1)
        bit_add(ind + 1, 1)
    return inv

INF = float('inf')
ans = INF
for odd_idxs in combinations(range(N), N//2):
    odd_idxset = set(odd_idxs)
    odds = []
    evens = []
    for i in range(N):
        if i in odd_idxset:
            odds.append((A[i] if i%2 else B[i], i))
        else:
            evens.append((B[i] if i%2 else A[i], i))
    odds.sort()
    evens.sort()
    vals = []
    inds = []
    for (e,ei),(o,oi) in zip(evens, odds):
        vals.append(e)
        vals.append(o)
        inds.append(ei)
        inds.append(oi)
    if len(evens) > len(odds):
        v,i = evens[-1]
        vals.append(v)
        inds.append(i)
    if any(a>b for a,b in zip(vals,vals[1:])): continue
    ans = min(ans, inversion(inds))

print(-1 if ans==INF else ans)