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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int main(void){

    int n;

    scanf("%d", &n);

    long long a[n];
    long long sum = 0, c = 0;

    for(int i = 0; i < n; i++){
        scanf("%lld", &a[i]);
    }

    long long b[n];

    for(int i = 0; i < n; i++){
        b[i] = a[i];
    }

    if(b[0] <= 0){
        c = 1 - b[0];
        b[0] = 1;
    }
    sum = b[0];

    for(int i = 1; i < n; i++){
        if(sum * (sum + b[i]) >= 0){
            if(sum > 0){
                c += b[i] + 1 + sum;
                b[i] = -1-sum;
            }else if(sum < 0){
                c += 1 - sum - b[i];
                b[i] = 1 - sum;
            }
        }
        sum += b[i];
    }

    long long d = 0;

    if(a[0] >= 0){
        d = a[0] + 1;
        a[0] = -1;
    }
    sum = a[0];

    for(int i = 1; i < n; i++){
        if(sum * (sum + a[i]) >= 0){
            if(sum > 0){
                d += a[i] + 1 + sum;
                a[i] = -1-sum;
            }else if(sum < 0){
                d += 1 - sum - a[i];
                a[i] = 1 - sum;
            }
        }
        sum += a[i];
    }

    if(c > d){
        printf("%lld\n", d);
    }else{
        printf("%lld\n", c);
    }

    return 0;
}