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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) > (b) ? (b) : (a))

int main(){
  int n;
  scanf("%d", &n);
  
  long long a[n];
  for(int i = 0; i < n; i++) scanf("%lld", a + i);
  
  long long initial_plus = max(-1 - a[0], 0);
  long long initial_minus = max(1 + a[0], 0);
  
  long long sum = 0;
  
  //初項を正にした場合
  sum = a[0] + initial_plus;
  for(int i = 1; i < n; i++){
    if((sum + a[i]) * sum >= 0ll){
      if(sum < 0ll){
        initial_plus += 1 - sum - a[i];
        sum = 1;
      }else{
        initial_plus += sum + a[i] + 1;
        sum = -1;
      }
    }else{
      sum += a[i];
    }
  }
  
  //初項を負にした場合
  sum = a[0] - initial_minus;
  for(int i = 1; i < n; i++){
    if((sum + a[i]) * sum >= 0ll){
      if(sum < 0ll){
        initial_minus += 1 - sum - a[i];
        sum = 1;
      }else{
        initial_minus += sum + a[i] + 1;
        sum = -1;
      }
    }else{
      sum += a[i];
    }
  }
  
  printf("%lld\n", min(initial_plus, initial_minus));
  
  return 0;
}