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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main()
{
    long long int n,i,a[100000],sum,count=0,flag=0;
    scanf("%lld",&n);
    for(i=0;i<n;i++){
        scanf("%lld",&a[i]);
        if(a[0]==0&&a[i]!=0&&flag==0) flag=i;
    }
    if(flag!=0){
        if( (a[flag]>0 && flag%2==0) || (a[flag]<0 && flag%2==1)) a[0]=1;
        else a[0]=-1;
        count++;
    }
    for(i=0;i<n;i++){
        if(i==0) sum=a[0];
        else{
            if(sum>0 && sum+a[i]>=0){
                count += 1+sum+a[i];
                a[i] = -1 * sum - 1;
                sum = -1;
            }
            else if(sum<0 && sum+a[i]<=0){
                count += 1-sum-a[i];
                a[i] = -1 * sum + 1;
                sum=1;
            }
            else if(sum+a[i]==0){
                if(sum>0) a[i]--;
                else if(sum<0) a[i]++;
                count++;
                sum+=a[i];
            }
            else sum += a[i];
        }
    }
    printf("%lld\n",count);
    return 0;
}