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 <cstdio>
#include <iostream>
using namespace std;
long long a [1000000];


long long min(long long a, long long b){
	int t = a;
	if(b <= t) t = b;
	return t;

}

int main(){
	int n;
	cin >> n;
	
	for(int t = 0; t < n; t++) cin >> a[t];
	

	long long sum = 0LL;
	long long x = 0LL;


	
		
		
		for(int t = 0; t < n; t++){
			
			sum += a[t];
			if(t % 2 == 1 && sum >= 0){
				long long s = sum + 1;
				sum = -1;
				x += s;
			}
			else if(t % 2 == 0 && sum <= 0){
				long long s = 1 - sum ;
				sum = 1;
				x += s;
			}
		}


		long long positive_x = x;
		x = 0LL;
		sum = 0LL;

	
		
		for(int t = 0; t < n; t++){
			
			
			sum += a[t];
			if(t % 2 == 0 && sum >= 0){
				long long s = sum + 1;
				sum = -1;
				x += s;
			}
			else if(t % 2 == 1 && sum <= 0){
				long long s = 1 - sum ;
				sum = 1;
				x += s;
			}
		}

		long long negative_x = x;
	
		long long result = min(positive_x, negative_x);

	    cout << result << endl;
}