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>
#include<ctype.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>


int main(){
	int n;
	scanf("%d",&n);
	long long a[n];
	int i;
	for(i=0;i<n;i++)
		scanf("%lld",&a[i]);
	
	long long sum[n];
	int j;
	sum[0]=a[0];
	for(i=0;i<n-1;i++){
		sum[i+1]=sum[i]+a[i+1];
	}
		
	long long ans=0;
	for(i=1;i<n;i++){
		if(sum[i]>=0 && sum[i-1]>0){
			ans+=(sum[i]+1);
			a[i]=-1-sum[i-1];
			for(j=i+1;j<n;j++)
				sum[j]-=(sum[i]+1);
			sum[i]=-1;
		}
		
		else if(sum[i]<=0 && sum[i-1]<0){
			ans+=(-sum[i]+1);
			a[i]=(-sum[i-1]+1);
			for(j=i+1;j<n;j++)
				sum[j]+=(-sum[i]+1);
			sum[i]=1;
		}

	}
	printf("%lld\n",ans);
	return 0;
}