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: Accepted)

#include<iostream>
using namespace std;

int main() {
  int N;
  cin >> N;

  int a[N];
  for (int i = 0; i < N; i++) cin >> a[i];

  long long odd_cnt = 0, odd_sum_cnt = 0;
  for (int i = 0; i < N; i++) {
    odd_sum_cnt += a[i];
    if (i % 2 == 0) {
      while (0 <= odd_sum_cnt) {
        odd_sum_cnt--;
        odd_cnt++;
      }
    } else {
      while (odd_sum_cnt <= 0) {
        odd_sum_cnt++;
        odd_cnt++;
      }
    }
  }

  long long even_cnt = 0, even_sum_cnt = 0;
  for (int i = 0; i < N; i++) {
    even_sum_cnt += a[i];
    if (i % 2 == 0) {
      while (even_sum_cnt <= 0) {
        even_sum_cnt++;
        even_cnt++;
      }
    } else {
      while (0 <= even_sum_cnt) {
        even_sum_cnt--;
        even_cnt++;
      }
    }
  }

  cout << min(odd_cnt, even_cnt) << endl;
  return 0;
}