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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n =sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        int[] sum = new int[n+1];
        for (int i = 1; i <= n; i++) {
            sum[i] = sum[i-1] + a[i-1];
        }

        int count1 = count(n, true, sum, 0 );
        int count2 = count(n, false, sum, 0);

        System.out.println(Math.min(count1, count2));
    }

    private static int count(int n , boolean pastPlus, int[] sum, long carry){
        int count2 = 0;
        for (int i = 1; i <= n; i++) {
            long cur = sum[i] + carry;

            if (pastPlus && cur >= 0) {
                // minus nisinaito
                count2 += cur + 1;
                carry = - cur - 1;
            }
            if (!pastPlus && cur <= 0) {
                count2 += -cur + 1;
                carry = -cur + 1;
            }

            pastPlus = !pastPlus;
        }

        return count2;
    }
}