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 chain
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))

def count_bit(x):
    res = 0
    while x:
        res += x % 2
        x >>= 1
    return res

def int2bin(x,N): 
    res = [0]*N
    for i in range(N):
        res[N-1-i] = x % 2
        x >>= 1
    return res

inf = 10**10
M = N//2 # the n of odd index
ans = inf
for b in range(2**N):
    valid = True 
    if count_bit(b) == M:
        odd_index = int2bin(b,N) # 1 : odd
        evens = []
        odds = []
        for i in range(N):
            if odd_index[i]:
                if i % 2:
                    odds.append((A[i],i))
                else:
                    odds.append((B[i],i))
            else:
                if i % 2:
                    evens.append((B[i],i))
                else:
                    evens.append((A[i],i))
        SE = sorted(evens)
        SO = sorted(odds)
        L = [None]*N
        L[::2] = SE
        L[1::2] = SO
        for (a,_),(b,_) in zip(L,L[1:]):
            if a > b:
                valid = False
                break
        if valid:
            L = list(zip(L,range(N)))
            L = sorted(L,key=lambda x : x[0][1])
            L = [(a,c) for (a,b),c in L]
            temp = 0
            for i in range(N):
                for j in range(i):
                    if L[j] > L[i]:
                        temp += 1
            ans = min(temp,ans)

if ans == inf:
    ans = -1
print(ans)



