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<string.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define ll long long

int main(){
	ll n,ai, sum = 0, ans = 0;
	int sign;

	scanf("%lld",&n);
	FOR(i,0,n){

		scanf("%lld",&ai);

		if(i == 0){
			if(ai >= 0) sign = 1;
			if(ai < 0) sign = -1;
		}

		sum += ai;

		//sumが正になるべきとき
		if((i%2 == 0 && sign == 1) || (i%2 == 1 && sign == -1)){
			if(sum <= 0){
				ans += -sum + 1;
				sum = 1;
			}
		}

		//sumが負になるべきとき
		if((i%2 == 1 && sign == 1) || (i%2 == 0 && sign == -1)){
			if(sum >= 0){
				ans += sum +1;
				sum = -1;
			}
		}

	}

	printf("%lld\n",ans);
}