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 Ruby (2.3.3) (judge verdict: Wrong Answer)

def sign(i)
  return 1 if i.positive? || i == 0
  -1
end

n = gets.chomp.to_i
a = gets.chomp.split.map { |item| item.to_i }
b = []
b[0] = a[0]

count = 0

if a[0] == 0
  case sign(a[1])
  when 1
    a[0] = -1
    count += 1
  when 0
    a[0] = 1
    count += 1
  end
end

(n-1).times do |i|
  b[i+1] = b[i] + a[i+1]
  if b[i+1] == 0
    if b[i] > 0
      a[i+1] -= 1
      count += 1
    else
      a[i+1] += 1
      count += 1
    end
  elsif sign(b[i]) == sign(b[i+1])
    if b[i].positive?
      count += (a[i+1] - (-b[i] - 1)).abs
      a[i+1] = -b[i] - 1
      b[i+1] = b[i] + a[i+1]
    else
      count += (a[i+1] - (-b[i] + 1)).abs
      a[i+1] = -b[i] + 1
      b[i+1] = b[i] + a[i+1]
    end
  end
end

puts count