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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
using namespace std;
int main(){
        int n;
        int N[100000];
        cin>>n;
        for(int i=0;i<n;i++){
                cin>>N[i];
        }
        for(int i=1;i<n;i++){
                N[i]=N[i]+N[i-1];
        }
        int ans=0; 
        if(N[0]==0){
                if(N[1]<=0){
                       for(int i=0;i<n;i++){
                               N[i]++;
                       ans=1;
                       }
                }
                else{
                        for(int i =0;i <n;i ++){
                                N[i]=N[i]-1;
                        }
                        ans=1;
                }
        }
        for(int i=1;i<n;i++){
                if(N[i-1]<0){
                    while(N[i]<=0){
                        for(int j=i;j<n;j++){
                                N[j]++;
                        }
                        ans++;
                    }
                }
                if(N[i-1]>0){
                    while(N[i]>=0){
                        for(int j=i;j<n;j++){
                                N[j]=N[j]-1;
                        }
                        ans++;
                    }
                }
        }
        cout<<ans;
}