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 <bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin >> N;
  vector<int> data(N);
  for(int i = 0; i < N; i++){
    cin >> data[i];
  }
  int count_odd, count_even;
  count_odd = 0;
  count_even = 0;
  int sum_odd = 0;
  int sum_even = 0;
  for(int i = 0; i < N; i++){
    sum_even += data[i];
    if(i%2 == 0){
      if(sum_even >= 0){
        count_even += (sum_even + 1);
        sum_even = -1;
      }
    }
    else{
      if(sum_even <= 0){
        count_even -= (sum_even - 1);
        sum_even = 1;
      }
    }
  }
  for(int i = 0; i < N; i++){
    sum_odd += data[i];
    if(i%2 == 0){
      if(sum_odd <= 0){
        count_odd -= (sum_odd - 1);
        sum_odd = 1;
      }
    }
    else{
      if(sum_odd >= 0){
        count_odd += (sum_odd + 1);
        sum_odd =1;
      }
    }
  }
  cout << min(count_even, count_odd)  << endl;
}