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>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

int main(){
    int n;
    scanf("%d", &n);
    lld a[n], b[n];
    rep(i, 0, n){
        scanf("%lld", &a[i]);
        b[i] = a[i];
    }
    lld sum_a = 0, sum_b = 0;
    lld cost_a = 0, cost_b = 0;
    rep(i, 0, n){
        sum_a += a[i];
        sum_b += b[i];
        if(i%2){
            if(sum_a >= 0){
                cost_a += -1*(-1 - sum_a);
                sum_a = -1;
            }
            if(sum_b <= 0){
                cost_b += 1 - sum_b;
                sum_b = 1;
            }
        }else{
            if(sum_a <= 0){
                cost_a += 1 - sum_a;
                sum_a = 1;
            }
            if(sum_b >= 0){
                cost_b += -1*(-1 - sum_b);
                sum_b = -1;
            }
        }
    }
    printf("%lld", cost_a<cost_b ? cost_a : cost_b);

    return 0;
}