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.Scanner;
import java.util.Arrays;
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];
        sum[0] = a[0];
        int count = 0;
        if(sum[0] == 0){
            count++;
        }
        count = solve(sum,a,count);
        System.out.println(count);
    }
    public static int solve(int[] sum,int[] a,int count){
        for(int i = 0;i < sum.length-1;i++){
            sum[i+1] = sum[i] + a[i+1];
            if(sum[0] > 0){
                if((i+1) % 2 == 1){
                    while(sum[i+1] >= 0){
                        sum[i+1]--;
                        count++;
                    }
                }
                if((i+1) % 2 == 0){
                    while(sum[i+1] <= 0){
                        sum[i+1]++;
                        count++;
                    }
                }
            }
            if(sum[0] < 0){
                if((i+1) % 2 == 1){
                    while(sum[i+1] <= 0){
                        sum[i+1]++;
                        count++;
                    }
                }
                if((i+1) % 2 == 0){
                    while(sum[i+1] >= 0){
                        sum[i+1]--;
                        count++;
                    }
                }
            }
        }
        return count;
    }
}