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>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
#define MAX 100005
int num[MAX];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&num[i]);
    ll sum,ans,cnt;
    sum=0;
    ans=0;
    for(int i=1;i<=n;i++)
    {
        sum+=num[i];
        if(sum<=0&&i%2)
        {
            ans+=1-sum;
            sum=1;
        }
        else if(sum>=0&&!(i%2))
        {
            ans+=sum+1;
            sum=-1;
        }
    }
    sum=0;
    cnt=0;
    for(int i=1;i<=n;i++)
    {
        sum+=num[i];
        if(sum>=0&&i%2)
        {
            cnt+=sum+1;
            sum=-1;
        }
        else if(sum<=0&&!(i%2))
        {
            cnt+=1-sum;
            sum=1;
        }
    }
    printf("%lld\n",min(ans,cnt));
    return 0;
}