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>
long long int min(long long int x, long long int y){return x<y?x:y;}
int main(void){
	int n,i;
	scanf("%d",&n);
	long long a,sum=0,osum=0,esum=0,o=0,e=0;
	for(i=0;i<n;i++){
		scanf("%lld",&a);
		osum+=a;
		esum+=a;
		if(i%2==0){
			if(osum<=0){
				o+=1-osum;
				osum=1;
			}
			if(esum>=0){
				e+=1+esum;
				esum=-1;
			}
		}else{
			if(osum>=0){
				o+=1+osum;
				osum=-1;
			}
			if(esum<=0){
				e+=1-esum;
				esum=1;
			}
		}
	}
	o = min(o,e);
	printf("%lld",o);
	return 0;
}