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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main(void){
    int n;
    scanf("%d",&n);
    long ar[n];
    for(int i = 0;i < n;i++)scanf("%ld",&ar[i]);
    long long cost1 = 0,cost2 = 0,difsum1 = 0,difsum2 = 0;
    long long cus_ar1[n];
    cus_ar1[0] = ar[0];
    for(int i = 1;i < n;i++)cus_ar1[i] = cus_ar1[i-1] + ar[i];
    long long cus_ar2[n];
    for(int i = 0;i < n;i++)cus_ar2[i] = cus_ar1[i];
    for(int i = 0;i < n;i++){
        cus_ar1[i] += difsum1;
        cus_ar2[i] += difsum2;
        if(i % 2){
            if(cus_ar1[i] >= 0){
                cost1 += cus_ar1[i] + 1;
                difsum1 -= cus_ar1[i] + 1;
            }
            if(cus_ar2[i] <= 0){
                cost2 += (-cus_ar2[i] + 1);
                difsum2 += (-cus_ar2[i] + 1);
            }
        }else{
            if(cus_ar1[i] <= 0){
                cost1 += (-cus_ar1[i] + 1);
                difsum1 += (-cus_ar1[i] + 1);
            }
            if(cus_ar2[i] >= 0){
                cost2 += cus_ar2[i] + 1;
                difsum2 -= cus_ar2[i] + 1;
            }
        }
    }
    printf("%lld",(cost1<cost2?cost1:cost2));
    return 0;
}