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{
	
	static final Scanner s=new Scanner(System.in);
	
	public static void main(String[] __){
		int n=s.nextInt();
		long a[]=new long[n];
		for(int i=0;i<n;i++) a[i]=s.nextInt();
		
		long sum[]=new long[n];
		sum[0]=a[0];
		
		long res=0;
		
		for(int i=1;i<n;i++) {
			sum[i]=a[i]+sum[i-1];
			if(sum[i-1]>0) {
				long t=-sum[i-1]-1;
				if(a[i]>t){
					res+=a[i]-t;
					a[i]=t;
				}
			}else {
				long t=-sum[i-1]+1;
				if(a[i]<t){
					res+=t-a[i];
					a[i]=t;
				}
			}
			sum[i]=a[i]+sum[i-1];
			
			//System.out.println(sum[i]);
		}
		
		//System.out.println(Arrays.toString(a));
		System.out.println(res);
	}
	
	
	static final boolean checkRev(long a,long b) {
		return (a<0&0<b)||(b<0&&0<a);
	}
}