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>
int A[100002];
int main() {
	long long N;
	scanf("%lld", &N);
	long long suma = 0, sumb = 0;
	long long  tmpa = 0, tmpb = 0;
	long long  ans = 0;
	//for (int i = 0; i < N; i++)
	for (long long i = 0; i < N; i++)
	{
		scanf("%d", &A[i]);

		if (i == 0)
		{
			suma += A[i];
			sumb += A[i];
		}
		//if (A[0] > 0)
		//{
		if (i > 0 && i % 2 == 1)
		{
			if (suma + A[i] >= 0)
			{
				tmpa += (suma + A[i] + 1);
				suma = -1;
			}
			else
			{
				suma = suma + A[i];
			}
		}
		if (i > 0 && i % 2 == 0)
		{
			if (suma + A[i] <= 0)
			{
				tmpa += (1 - (suma + A[i]));
				suma = 1;
			}
			else
			{
				suma = suma + A[i];
			}
		}
		//}

		//if (A[0] < 0)
		//{
		if (i > 0 && i % 2 == 1)
		{
			if (sumb + A[i] <= 0)
			{
				tmpb += (1 - (sumb + A[i]));
				sumb = 1;
			}
			else
			{
				sumb = sumb + A[i];
			}
		}
		if (i > 0 && i % 2 == 0)
		{
			if (sumb + A[i] >= 0)
			{
				tmpb += ((sumb + A[i] + 1));
				sumb = -1;
			}
			else
			{
				sumb = sumb + A[i];
			}
		}
		//}


	}
	if (tmpa < tmpb)
	{
		ans = tmpa;
	}
	else
	{
		ans = tmpb;
	}

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