1206_B. Make Product Equal One
You are given n numbers a_1, a_2, ..., 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 ⋅ a_2 ... ⋅ 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⋅ (-1) ⋅ (-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 ≤ n ≤ 10^5) — the number of numbers.

The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 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.
```python
def func(lst):
    prod = 1
    for item in lst:
        if item!=0:
            prod = prod*item
    return prod
    
w = int(input())
a=list(map(int,input().split()))

    
count=0
count1=0
 
for i,item in enumerate(a):
    if item<0 and item !=-1:
        count+=abs(item+1)
        a[i]=-1
        
    if item>0 and item!=1:
        count+=item-1
        a[i]=1
        
    if item ==0:
        count1 += 1
        

prod = func(a)
if prod == -1 and count1==0:
    count = count+2
else:
    count = count+ count1


    
print(count)
```