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: Wrong Answer)

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;

int main(){
    //freopen("testcase", "r", stdin);
    int N, temp;
    vector<int> a;
    scanf("%d", &N);
    int start = 0;
    bool v = false;
    for(int i=0; i<N; i++){
        scanf("%d", &temp);
        if (temp==0){
            if(!v){
                start+=1;
            }
        }else if(!v) v=true;
        a.push_back(temp);
    }

    long long int sum=0, cnt=0;
    if(start!=0){
        cnt = 2*(start-1)+1;
        if(a[start]>0){
            if(a[start]>1){
                sum=a[start]-1;
            }else{
                sum=1;
                cnt+=1;
            }
        }else{
            if(a[start]<-1){
                sum=a[start]+1;
            }else{
                sum=-1;
                cnt+=1;
            }
        }
    }else{
        sum=a[start];
    }
    start++;
    for(size_t i = start; i!=a.size(); i++){
        if(sum+a[i]==0){
            if(sum>0){
                sum=-1;
            }else{
                sum=1;
            }
            cnt+=1;
            continue;
        }
        if(sum+a[i]>0 && sum>0){
            cnt += sum+a[i]+1;
            sum=-1;
        }else if(sum+a[i]<0 && sum<0){
            cnt += 1-sum-a[i];
            sum=1;
        }else{
            sum+=a[i];
        }
    }
    if(sum==0) cnt+=1;
    printf("%lld\n", cnt);



    return 0;

}