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 double sequence(int a[], double start) {
		double count = 0.0, presum = -1.0 * start, sum = 0.0;
		
		for(int i : a) {
			sum += (double)i;
			if(i == 0)sum += start;
			if(sum * presum > 0) {
				double min = Math.abs(sum) + 1;
				if(presum > 0)sum -= min;
				else sum += min;
				count += min;
			}
			
			if(sum == 0) {
				if(presum > 0)sum--;
				else sum++;
				++count;
			}
			
			presum = sum;
		}
		
		return count;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n, a[];
		double count = 0;
		
		n = sc.nextInt();
		a = new int[n];
		
		for(int i = 0; i < n; ++i) a[i] = sc.nextInt();
		
		sc.close();
		
		if(a[0] == 0)a[0]++;
		
		int tmp = Math.abs(a[0]) + 1;
		
		if(a[0] > 0)tmp = a[0] - tmp;
		else tmp = a[0] + tmp;
		
		count = Math.min(sequence(a, (double)a[0]),sequence(a, (double)tmp));
		
		System.out.printf("%.0f\n", count);
	}

}