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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) throws Exception {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[] as = new int[n];
    for (int i = 0; i < n; i++) {
      as[i] = sc.nextInt();
    }
    long[] sums = new long[n];

    sums[0] = as[0];
    for (int i = 1; i < n; i++) {
      sums[i] = sums[i - 1] + as[i];
    }
    long count = 0;
    for (int i = 1; i < n; i++) {
      long prev = sums[i - 1];
      long now = sums[i];
      if (prev < 0 && 0 < now || 0 < prev && now < 0) {
        continue;
      }
      long add = -now + (now < 0 ? 1 : -1);
      for (int j = i; j < n; j++) {
        sums[j] += add;
      }
      count += Math.abs(add);
    }
    System.out.println(count);
  }
}