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: Accepted)

#include<iostream>
using namespace std;
long long a[100005],b[100005];
int main(){
    long long n,result=0,t=0;
    cin>>n>>b[0];
    for(int i=1;i<n;i++){
        long long l;
        cin>>l;
        b[i]=b[i-1]+l;
    }
    for(int i=0;i<n;i++)a[i]=b[i];
    long long result0;
    if(a[0]>0){
        result0=0;
        t=0;
    }else if(a[0]<=0){
        result0=1-a[0];
        t=1-a[0];
        a[0]=1;
    }
    for(int i=1;i<n;i++){
        a[i]=a[i]+t;
        if(a[i-1]>0&&a[i]>=0){
            result0+=a[i]+1;
            t-=a[i]+1;
            a[i]=-1;
        }
        else if(a[i-1]<0&&a[i]<=0){
            result0+=1-a[i];
            t+=1-a[i];
            a[i]=1;
        }
    }
        for(int i=0;i<n;i++)a[i]=b[i];
        if(a[0]<0){
            result=0;
            t=0;
        }else if(a[0]>=0){
            result=1+a[0];
            t=-1-a[0];
            a[0]=-1;
        }
        for(int i=1;i<n;i++){
            a[i]=a[i]+t;
        if(a[i-1]>0&&a[i]>=0){
            result+=a[i]+1;
            t-=a[i]+1;
            a[i]=-1;
        }
        else if(a[i-1]<0&&a[i]<=0){
            result+=1-a[i];
            t+=1-a[i];
            a[i]=1;
        }
        }
        cout<<((result<result0)?result:result0)<<endl;
    return 0;
}