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 <string.h>

int abs(int x)
{
	if(x < 0)
		return -x;
	else
		return x;
}

int main(void)
{
	int n;
	int a[100000];
	int i, j;
	int sum = 0;
	int ans = 0;
	char sign;
	
	/* ----- input ----- */
	scanf("%d", &n);
	for(i=0; i<n; i++)
		scanf("%d", &a[i]);
	
	/* ----- to Do ----- */
	for(i=0; i<n; i++){
		if(a[i] > 0){
			sign = 'p';
			break;
		}
		else if(a[i] < 0){
			sign = 'm';
			break;
		}
		else
			continue;
	}
	for(; i<n; i++){
		sum += a[i];
		
		if(sum == 0){
			for(j=i+1; j<n; j++){
				if(sum == a[j])
					continue;
				else if(sign == 'p'){
					sum++;
					ans++;
					i = j;
					break;
				}
				else if(sign == 'm'){
					sum--;
					ans++;
					i = j;
					break;
				}
			}
			i = j;
			ans++;
		}
		else if(sum > 0 && sign == 'p')
			sign = 'm';
		else if(sum < 0 && sign == 'm')
			sign = 'p';
		else{
			if(sign == 'p'){
				ans += abs(sum)+1;
				sum += abs(sum)+1;
				sign = 'm';
			}
			else if(sign == 'm'){
				ans += abs(sum)+1;
				sum -= abs(sum)+1;
				sign = 'p';
			}
		}
	}
	
	/* ------ output ----- */
	printf("%d\n", ans);
	
	return 0;
}