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<string>
using namespace std;
typedef long long ll;
long long a[100000], b[100001];
int main() {
	int long long n;
	cin >> n;
	for (ll i = 0; i < n; i++) {
		cin >> a[i];
		b[i] += a[i];
		b[i + 1] = b[i];
	}
	int long long sum = 0, sum2 = 0;
	if (b[0] == 0) {
		ll i = 0;
		while (b[i] == 0&&i<n)
			i++;
		if (i % 2 == 0) {
			b[0] = 1;
			sum2 = 1;
		}
		else {
			b[0] = -1;
			sum2 = -1;
		}
		sum++;
	}
	for (ll i = 1; i < n; i++) {
		b[i] += sum2;
		if (b[i] == 0) {
			if (b[i - 1] > 0) {
				sum2--;
				sum++;
				b[i] = -1;
			}
			else {
				sum2++;
				sum++;
				b[i] = 1;
			}
		}
		else {
			if (b[i - 1] > 0 && b[i] > 0) {
				sum2 -= (b[i]+1);
				sum += (b[i] + 1);
				b[i] = -1;
			}
			else if (b[i] < 0 && b[i - 1] < 0) {
				sum2 += (0 - b[i] + 1);
				sum += (0 - b[i] + 1);
				b[i] = 1;
			}
		}
	}
	cout << sum << endl;
}