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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
 
public class Main {
 
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] a = new int[n];
		int[] s = new int[n];
		long sum = 0;
		a[0] = scanner.nextInt();
		sum += a[0];
 
		boolean check;
		if(a[0] < 0){
			check = false;
		}else{
			check = true;
		}
		long count = 0;
		for(int i=1;i<n;i++){
			a[i] = scanner.nextInt();
			long x = sum + a[i];
			long y = 0;
			if(check && x >= 0){
				y = -1 - x;
			}else if(!check && x < 0){
				y = 1 - x;
			}else if(!check && x == 0){
				y = 1;
			}
 
			a[i] += y;
			count += Math.abs(y);
			sum += a[i];
			//System.out.println(y + " " + a[i] + " " + sum);
			check = !check;
		}
		if(sum == 0){
			count ++;
		}
		System.out.println(count);
 
	}
 
}