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
# coding: utf-8
import sys
from itertools import permutations,combinations
sys.setrecursionlimit(10 ** 7)

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

# 全探索
INF = 10 ** 5
answer = INF
N = ir()
if N == 1:
    print(0); exit()
A = lr()
B = lr()
for pattern in combinations(range(N), (N+1)//2):
    pattern = set(pattern)
    red = []
    blue = []
    for i in range(N):
        if i in pattern:
            red.append((B[i], i) if i&1 else (A[i], i))
        else:
            blue.append((A[i], i) if i&1 else (B[i], i))
    red.sort()
    blue.sort()
    result = []
    
    for i in range(N//2):
        if red[i][0] > blue[i][0]:
            break
        if i + 1 < len(red) and red[i+1][0] < blue[i][0]:
            break
        result.append(red[i][1])
        result.append(blue[i][1])
    else:
        if N&1 and blue[-1][0] > red[-1][0]:
            continue
        if N&1:
            result.append(red[-1][1])
        temp = 0
        used = set()
        for i in range(N):
            for x in used:
                if result[i] < x:
                    temp += 1
            used.add(result[i])            
        if temp < answer:
            answer = temp

print(answer if answer != INF else -1)
# 12