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 {
    static long ans;
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int [] num = new int [N];
        long sum = 0;
        ans=0;
        for(int i=0; i<N; i++){
            num[i] = sc.nextInt();
            sum+=num[i];
            if(i!=0)sum=function(i,num,sum);
        }
        System.out.println(ans);
    }
    static long function(int i, int[]num, long sum){
        int a = sign((sum-num[i]));
        int b = sign(sum);
        int t = a*b;
        if(t==0){
            ans++;
            if(a>0){
                return -1;
            }else{
                return 1;
            }
        }else if(t<0){
            return sum;
        }else{
            ans+=(Math.abs(sum)+1);
            if(sum>0){
                return -1;
            }else{
                return 1;
            }
        }
    }
    static int sign(long A){
        if(A>0){
            return 1;
        }else if(A<0){
            return -1;
        }else{
            return 0;
        }
    }
}