Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length N. The i-th term in the sequence is a_i.
In one operation, you can select a term and either increment or decrement it by one.

At least how many operations are necessary to satisfy the following conditions?

For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero.

For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.

Constraints

2 ≤ n ≤ 10^5

|a_i| ≤ 10^9

Each a_i is an integer.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print the minimum necessary count of operations.

Sample Input 1

4
1 -3 1 0

Sample Output 1

4

For example, the given sequence can be transformed into 1, -2, 2, -2 by four operations. The sums of the first one, two, three and four terms are 1, -1, 1 and -1, respectively, which satisfy the conditions.

Sample Input 2

5
3 -6 4 -5 7

Sample Output 2

0

The given sequence already satisfies the conditions.

Sample Input 3

6
-1 4 3 2 -5 4

Sample Output 3

8

Solution in Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)

function main()
	n = readline() |> parseInt
	a = map(parseInt, split(readline()))
	b = Array{Int}(n)
	k = 0
	if a[1] > 0
		b[1] = 1
		k += abs(a[1]+1)
	else
		b[1] = -1
		k += abs(a[1]-1)
	end
	for i in 2:n
		b[i] = a[i]+b[i-1]
		if b[i]*b[i-1] >= 0
			if b[i-1] < 0
				k += abs(b[i]-1)
       			b[i] = 1
			else
				k += abs(b[i]+1)
				b[i] = -1
			end
		end
	end
	c = Array{Int}(n)
	l = 0
	if a[1] > 0
		c[1] = -1
		l += abs(a[1]+1)
	else
		c[1] = 1
		l += abs(a[1]-1)
	end
	for i in 2:n
		c[i] = a[i]+c[i-1]
		if c[i]*c[i-1] >= 0
			if c[i-1] < 0
				l += abs(c[i]-1)
       			c[i] = 1
			else
				l += abs(c[i]+1)
				c[i] = -1
			end
		end
	end
	print(min(k,l))
end

main()