I found an interesting problem on https://codeforces.com/problemset/problem/1206/B:
You are given $n$ numbers $a_1, a_2, \dots, a_n$. With a cost of one coin you can perform the following operation:

Choose one of these numbers and add or subtract $1$ from it.

In particular, we can apply this operation to the same number several times.

We want to make the product of all these numbers equal to $1$, in other words, we want $a_1 \cdot a_2$ $\dots$ $\cdot a_n = 1$. 

For example, for $n = 3$ and numbers $[1, -3, 0]$ we can make product equal to $1$ in $3$ coins: add $1$ to second element, add $1$ to second element again, subtract $1$ from third element, so that array becomes $[1, -1, -1]$. And $1\cdot (-1) \cdot (-1) = 1$.

What is the minimum cost we will have to pay to do that?


-----Input-----

The first line contains a single integer $n$ ($1 \le n \le 10^5$) — the number of numbers.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10^9 \le a_i \le 10^9$) — the numbers.


-----Output-----

Output a single number — the minimal number of coins you need to pay to make the product equal to $1$.


-----Examples-----
Input
2
-1 1

Output
2
Input
4
0 0 0 0

Output
4
Input
5
-5 -3 5 3 0

Output
13


-----Note-----

In the first example, you can change $1$ to $-1$ or $-1$ to $1$ in $2$ coins.

In the second example, you have to apply at least $4$ operations for the product not to be $0$.

In the third example, you can change $-5$ to $-1$ in $4$ coins, $-3$ to $-1$ in $2$ coins, $5$ to $1$ in $4$ coins, $3$ to $1$ in $2$ coins, $0$ to $1$ in $1$ coin.

I tried it in Python, but could not do it. Can you solve it?
q=int(input())
w=list(map(int,input().split()))
e=0
r=0
t=0
for i in w:
    if i<0:
        e+=-1-i
        r+=1
    elif i>0:
        e+=i-1
    else:
        e+=1
        t+=1
if r%2==1:
    if t>0:
        print(e)
    else:
        print(e+2)
else:
    print(e)