Problem: 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

Solution (in PYTHON3):

Solution:
import itertools
import math
import os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353

# 2**18 == 262144
N = int(sys.stdin.buffer.readline())
A = list(map(int, sys.stdin.buffer.readline().split()))
B = list(map(int, sys.stdin.buffer.readline().split()))


class BinaryIndexedTree:
    # http://hos.ac/slides/20140319_bit.pdf
    def __init__(self, size):
        """
        :param int size:
        """
        self._bit = [0] * size
        self._size = size

    def add(self, i, w):
        """
        i 番目に w を加える
        :param int i:
        :param int w:
        """
        x = i + 1
        while x <= self._size:
            self._bit[x - 1] += w
            x += x & -x

    def sum(self, i):
        """
        [0, i) の合計
        :param int i:
        """
        ret = 0
        while i > 0:
            ret += self._bit[i - 1]
            i -= i & -i
        return ret

    def __len__(self):
        return self._size


# 18C9 < 50000
# どれを偶数番目にするか決まれば回数も分かる
ans = INF
for evens_i in itertools.combinations(range(N), r=(N + 1) // 2):
    is_even = [False] * N
    for i in evens_i:
        is_even[i] = True
    evens = []
    odds = []
    for i, even in enumerate(is_even):
        if even:
            evens.append((A[i] if i % 2 == 0 else B[i], i))
        else:
            odds.append((A[i] if i % 2 == 1 else B[i], i))
    evens.sort()
    odds.sort()

    prev = -1
    # 転倒数
    bit = BinaryIndexedTree(size=N)
    ok = True
    cnt = 0
    for i in range(N):
        c, mi = [evens, odds][i % 2][i // 2]
        if c < prev:
            ok = False
            break
        prev = c
        cnt += i - bit.sum(mi)
        bit.add(mi, 1)
    if ok:
        ans = min(ans, cnt)
        # print(evens_i, cnt, evens, odds)
if ans == INF:
    print(-1)
else:
    print(int(ans))