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: Accepted)

#include <stdio.h>
int main()
{
int i,n;
long long oddc=0,evenc=0;
long long odds=0,evens=0,a[100001];
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%lld",&a[i]);
odds+=a[i];
evens+=a[i];
  if(i%2==0){
    if(odds<=0){
    oddc=oddc+1-odds;
    odds=1;
    }
    if(evens>=0){
    evenc=evenc+1+evens;
    evens=-1;
    }
  }else{
    if(odds>=0){
    oddc=oddc+1+odds;
    odds=-1;
    }
    if(evens<=0){
    evenc=evenc+1-evens;
    evens=1;
    }
  }
}
if(oddc<evenc){
printf("%lld",oddc);
}else{
printf("%lld",evenc);
}
return 0;
}