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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
using namespace std;
long long  mod = 1000000007;
long long a[100010];
int main()
{
	int N;
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> a[i];
	}
	long long sum = 0, an = 0;
	sum = a[0];
	for (int i = 1; i < N; i++) {
		if (sum > 0) {
			if (sum + a[i] >= 0) {
				an += abs(a[i] - (-1 - sum));
				sum = -1;
				
			}
			else {
				sum += a[i];
			}
		}
		else {
			if (sum + a[i] <= 0) {
				an += abs(a[i] - (1 - sum));
				sum = 1;
			}
			else {
				sum += a[i];
			}
		}
	}
	sum = -a[0];
	long long an2 = abs(2 * a[0]);
	for (int i = 1; i < N; i++) {
		if (sum > 0) {
			if (sum + a[i] >= 0) {
				an2 += abs(a[i] - (-1 - sum));
				sum = -1;

			}
			else {
				sum += a[i];
			}
		}
		else {
			if (sum + a[i] <= 0) {
				an2 += abs(a[i] - (1 - sum));
				sum = 1;
			}
			else {
				sum += a[i];
			}
		}
	}
	cout << min(an, an2);
}