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 calcs(x,y,ax,ay,bx,by) ((ax-x)*(by-y)-(bx-x)*(ay-y))/2.0 end
def inpf() a=gets.chomp.split(" ").map(&:to_f)end
def inps() a=gets.chomp.split(" ")end  
def copy(a) Marshal.load(Marshal.dump(a)) end
def kaizyo(n)(n < 2)? 1 : (2..n).inject(:*) end
def scount(a) b=na(a.max+1);a.each{|n|b[n]+=1};return b end
def na(n=0,d=0) Array.new(n,d)end
def na2(n=0,m=0,d=0) Array.new(n){Array.new(m,d)}end
def na3(n=0,m=0,l=0,d=0) Array.new(n){Array.new(m){Array.new(l,d)}}end
def bit(n) n.to_s(2).split("").map(&:to_i) end
def inp() a=gets.chomp.split(" ").map(&:to_i)end
n = inp[0]
a=inp
sum = a[0]
summ = sum
c = 0
if(sum == 0)
  sum = -1
  c+=1
end
a.delete_at(0)
(a).each do |d|
  if(sum >0)
    if(sum+d < 0)
      sum+=d
    else
      c+=((sum+d)+1)
      sum = -1
    end
  else
    if(sum+d > 0)
      sum+=d
    else
      c+=(1-(sum+d))
      sum = 1
    end
  end
end
min = c
c = 0
sum = summ
if(sum == 0)
  sum = 1
  c+=1
end
(a).each do |d|
  if(sum >0)
    if(sum+d < 0)
      sum+=d
    else
      c+=((sum+d)+1)
      sum = -1
    end
  else
    if(sum+d > 0)
      sum+=d
    else
      c+=(1-(sum+d))
      sum = 1
    end
  end
end
p [c,min].min
=begin
=end