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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
int main(void)
{
    int i;
    int n;
    int s[10050] = {0};
    int count;//正か不か
    int wa = 0;
    int ans = 0;
    scanf("%d",&n);

    for(i=0;i<n;i++)
        scanf("%d",&s[i]);

    if(s[0] < 0){
        count = 1;//次が不
    }else{
        count = 0;
    }
    wa = s[0];
    for(i=1;i<n;i++){
        if((wa+s[i] == 0) && (count == 0)){
            count = 1;
            ans += 1;
            s[i]++;
            wa += s[i];
            //printf("ans == %d\n",ans);
        }else if((wa+s[i] > 0) && (count == 0)){
            count = 1;
            ans += abs(wa) +1-s[i];
            s[i] = abs(wa)-s[i];
            wa += s[i];
            //printf("ans == %d\n",ans);
           /* count = 1;
            wa += s[i];
            printf("ans == %d\n",ans);*/

        }else if((wa+s[i] < 0) && (count == 0)){
            count = 1;
            wa += s[i];
            printf("ans == %d\n",ans);
            /*count = 1;
            ans += -wa +1-s[i];
            s[i] = -wa + 1-s[i];
            wa += s[i];
            printf("ans == %d\n",ans);*/

        }else if((wa+s[i] < 0) && (count == 1)){
            /*count = 0;      
            wa = s[i];      
            printf("ans == %d\n",ans);*/
            count = 0;
            ans += abs(wa) - s[i] +1;
            s[i] = abs(wa) + s[i];
            wa += s[i];
            //printf("ans == %d %d\n",ans,s[i]);

        }else if((wa+s[i] == 0) && (count == 1)){
            count = 0;
            ans += 1;
            s[i]--;
            wa += s[i];
            printf("ans == %d\n",ans);
        }else if((wa+s[i] > 0) && (count == 1)){
            /*count = 0;
            ans += -wa - s[i] +1;
            s[i] = -(wa + s[i] +1);
            wa += s[i];
            printf("ans == %d\n",ans);*/
            count = 0;      
            wa = s[i];      
            //printf("ans == %d\n",ans);

        }
    }
    printf("%d\n",ans);
    return 0;
}