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

int in(void){
	int i;scanf("%d",&i);
	return i;
}
long long llin(void){
	long long i;scanf("%lld",&i);
	return i;
}
void chin(char s[]){
	scanf("%s",s);
}
void print(int a){
	printf("%d\n",a);
}
void llprint(long long a){
	printf("%lld\n",a);
}
void print2(int a,int b){
	printf("%d %d\n",a,b);
}
long long max(long long a,long long b){
	return a>b?a:b;
}
long long min(long long a,long long b){
	return a<b?a:b;
}

int main(void){
	long long n=llin(),a[100000],i,ans=0,sum=0;
	for(i=0;i<n;i++){
		a[i]=llin();
	}
	for(i=0;i<n;i++){
		sum+=a[i];
		if(sum==0){
			if(sum-a[i]<0){
				sum=1;
				ans++;
			}
			else if(sum-a[i]>0){
				sum=-1;
				ans++;
			}
			else if(a[i+1]>0){
				sum=-1;
				ans++;
			}
			else{
				sum=1;
				ans++;
			}
		}
		else if(sum-a[i]<0&&sum<0){
			ans+=fabs(1-sum);
			sum=1;
		}
		else if(sum-a[i]>0&&sum>0){
			ans+=fabs(1+sum);
			sum=-1;
		}
	}
	llprint(ans);
	return 0;
}