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 <stdlib.h>
#include <stdio.h>

int main()
{   int n;
    scanf("%d",&n);
    long long a[n];
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
    }
    long long mans=0,pans=0;
    long long msub[n],psub[n];
    if(a[0]==0){
        pans=1;
        mans=1;
        psub[0]=1;
        msub[0]=-1;
    }
    if(a[0]>0){
        psub[0]=a[0];
        msub[0]=-1;
        mans=a[0]+1;
    }else{
        psub[0]=1;
        msub[0]=a[0];
        pans=a[0]+1;
    }
    int i;
    for(i=1;i<n;i++){
        if(i%2==1&&psub[i-1]+a[i]>=0){
            pans+=(psub[i-1]+a[i]+1);
            psub[i]=-1;
        }
        else if(i%2==0&&psub[i-1]+a[i]<=0){
            pans+=(1-(psub[i-1]+a[i]));
            psub[i]=1;
        }else{
            psub[i]=(psub[i-1]+a[i]);
        }
        if(i%2==1&&msub[i-1]+a[i]<=0){
            mans+=(1-(msub[i-1]+a[i]));
            msub[i]=1;
        }else if(i%2==0&&msub[i-1]+a[i]>=0){
            mans+=(msub[i-1]+a[i]+1);
            msub[i]=-1;
        }else{
            msub[i]=(msub[i-1]+a[i]);
        }
    }
    
    printf("%lld",mans<pans?mans:pans);

    return 0;
}