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>

int main(int argc, char *argv[])
{
	int i, n, sign;
	long long a, c = 0, t = 0;
	scanf("%d", &n);
	scanf("%lld", &t);
	sign = t>0?-1:1;
	for ( i = 1; i < n; i++ ){
		scanf("%lld", &a);
		if ( sign < 0 ){
			if ( t+a >= 0 ){
				c += llabs(-1-t-a);
				t = -1;
			} else {
				t += a;
			}
		} else {
			if ( t+a <= 0 ){
				c += llabs(1-t-a);
				t = 1;
			} else {
				t += a;
			}
		}
		sign *= -1;
	}
	printf("%lld\n", c);
	return 0;
}