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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main()
{
  int n;
  int a[100001];
  int i;
  int sum;
  int ans1=0,ans2=0;
  int ans;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&a[i]);
  }
  /*a[0]を正とするとき*/
  sum=a[0];
  if(a[0]<=0){
    ans1+=-a[0]+1;
    sum+=ans1;
  }
  for(i=1;i<n;i++){
    sum+=a[i];
    if(i%2==0){
      if(sum<=0){
        ans1+=-sum+1;
        sum+=ans1;
      }
    }
    else if(i%2==1){
      if(sum>=0){
        ans1+=sum+1;
        sum-=ans1;
      }
    }
  }
  /*a[0]を負とするとき*/
  sum=a[0];
  if(a[0]>=0){
    ans2+=a[0]+1;
    sum+=ans2;
  }
  for(i=1;i<n;i++){
    sum+=a[i];
    if(i%2==1){
      if(sum<=0){
        ans2+=-sum+1;
        sum+=ans2;
      }
    }
    else if(i%2==0){
      if(sum>=0){
        ans2+=sum+1;
        sum-=ans2;
      }
    }
  }
  if(ans1>=ans2) ans=ans2;
  else ans=ans1;
  printf("%d",ans);
  return 0;
}