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:
# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7

def pop_count(i):
    i = i - ((i >> 1) & 0x55555555)
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
    i = (i + (i >> 4)) & 0x0f0f0f0f
    i = i + (i >> 8)
    i = i + (i >> 16)
    return i & 0x3f

N = INT()
cards = [[]] * 2
cards[0] = LIST()
cards[1] = LIST()

ans = INF
for S in range(1<<N):
    # 無理なケース(奇数枚裏返す)を除外
    if pop_count(S) % 2 != 0:
        continue
    # 奇数位置と偶数位置で分けて保持する
    A1 = [[] for i in range(2)]
    for i in range(N):
        # カードiを青くする
        if S & 1<<i:
            A1[(i%2)^1].append((cards[1][i], i))
        # 赤のまま
        else:
            A1[i%2].append((cards[0][i], i))
    # 無理なケース(置きたい偶奇位置の数がおかしい)を除外
    if len(A1[0]) != ceil(N, 2) or len(A1[1]) != N // 2:
        continue
    # 偶奇が同じ位置同士は自由に動けるのでソートする
    A1[0].sort(reverse=1)
    A1[1].sort(reverse=1)
    cnt = 0
    A2 = [0] * N
    idxs = [0] * N
    for i in range(N):
        # 偶奇位置を交互に取り出す
        A2[i], idxs[i] = A1[i%2].pop()
    # ソート済と一致すればOK
    if A2 == sorted(A2):
        # 移動距離を転倒数数え上げで求める(Nが小さいのでN^2かけて大丈夫)
        cnt = 0
        for i, idx1 in enumerate(idxs):
            for j, idx2 in enumerate(idxs[i+1:], i+1):
                if idx1 > idx2:
                    cnt += 1
        ans = min(ans, cnt)
if ans == INF:
    print(-1)
else:
    print(ans)