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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int min(int a, int b){
	if(a < b){
		return a;
	}
	else{
		return b;
	}
}

int main(){
	int n, i;
	scanf("%d", &n);
	int *a = (int *)malloc(sizeof(int) * n);
	for(i = 0; i < n; i++){
		scanf("%d", &a[i]);
	}
	long long int sum = 0;
	int ans1 = 0, ans2 = 0, f = -1;
	for(i = 0; i < n; i++){
		sum += a[i];
		if(f == 1 && sum <= 0){
			ans1 += 1 - sum;
			sum = 1;
		}
		if(f == -1 && sum >= 0){
			ans1 += 1 + sum;
			sum = -1;
		}
		if(f == 1){
			f = -1;
		}
		else if(f == -1){
			f = 1;
		}
	}
	f = 1;
	sum = 0;
	for(i = 0; i < n; i++){
		sum += a[i];
		if(f == 1 && sum <= 0){
			ans2 += 1 - sum;
			sum = 1;
		}
		if(f == -1 && sum >= 0){
			ans2 += 1 + sum;
			sum = -1;
		}
		if(f == 1){
			f = -1;
		}
		else if(f == -1){
			f = 1;
		}
	}
	printf("%d\n", min(ans1, ans2));
	return 0;
}