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 Python (3.4.3) (judge verdict: Wrong Answer)

import copy
n = int(input())
a = [int(i) for i in input().split()]
b=a.copy()
s0p = a[0]
s0n = b[0]
countp = 0
countn = 0

if a.count(0)==n:
  print(2*n+1)
  exit()

if s0p<=0:
  s0p+=(abs(s0p)+1)
  countp+=1

if s0n>=0:
  s0n-=(abs(s0n)+1)
  countn+=1

for i in range(1,n):
  s1 = s0p+a[i]
  if s0p*s1>=0:
    if s1>0:
      a[i]-=(abs(s1)+1)
      countp+=(abs(s1)+1)
    elif s1<0:
      a[i]+=(abs(s1)+1)
      countp+=(abs(s1)+1)
    elif s1==0:
      if s0p>0:
        a[i]-=1
        countp+=1
      elif s0p<0:
        a[i]+=1
        countp+=1
  s0p += a[i]

for i in range(1,n):
  s1 = s0n+b[i]
  if s0n*s1>=0:
    if s1>0:
      b[i]-=(abs(s1)+1)
      countn+=(abs(s1)+1)
    elif s1<0:
      b[i]+=(abs(s1)+1)
      countn+=(abs(s1)+1)
    elif s1==0:
      if s0n>0:
        b[i]-=1
        countn+=1
      elif s0n<0:
        b[i]+=1
        countn+=1
  s0n += b[i]

print(countp if countp<=countn else(countn))