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 main()
{
	long long int n;
	scanf("%lld",&n);
	long long int i;
	long long int a[100005];
	for(i=0;i<n;i++)
		scanf("%lld",&a[i]);
	long long int b[100005];
	for(i=0;i<n;i++)
		b[i]=a[(i+1)%n]-a[i];
	long long int sum=0;
	for(i=0;i<n;i++)
		sum+=a[i];
	if(sum%(n*(n+1)/2)>0)
	{
		printf("NO\n");
		return 0;
	}
	sum/=n*(n+1)/2;
	for(i=0;i<n;i++)
		b[i]=sum-b[i];
	long long int cnt=0;
	for(i=0;i<n;i++)
	{
		if(b[i]%n>0)
		{
			printf("NO\n");
			return 0;
		}
		cnt+=b[i]/n;
	}
	if(cnt==n)
		printf("YES\n");
	else
		printf("NO\n");
	return 0;
}