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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int check(long long int a)
{
     if(a>0)
      return 1;
     if(a<0)
      return 2;
     return 0;
}
int main() 
{
    long long int sum,n,i,cnt=0;
    cin>>n;
    long long int x,a[n],h[n];
    for(i=0;i<n;i++)
     cin>>a[i];
    sum=a[0];
    if(sum==0)
     cnt++,sum++;
    h[0]=sum;
    for(i=1;i<n;i++)
    {
      if(sum==0)
      {
          cnt++;
          if(h[i-1]>0)
           h[i]=sum=1;
          else
           h[i]=sum=-1;
          sum=1;
          continue;
      }
      if(check(sum+a[i])==check(sum))
      {
          if(sum<0)
          {
               x=1-sum;
               cnt+=(x-a[i]);
               sum=1;
          }
          else
          {
            x=-1-sum;
            cnt+=(a[i]-x);
            sum=-1;
          }
      }
      else
       sum+=a[i];
     h[i]=sum;
    }
    cout<<cnt;
    return 0;
}