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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <string>
#include <algorithm>
#include <iostream>
#include <string>
#include <map>
#include <set>
#include <functional>
#include <iostream>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n;
ll a[1000001];
int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%lld",&a[i]);
	}
	ll res=0;
	bool flag=true;
	ll now=0;
	for(int i=0;i<n;i++){
		if(now==0){
			now+=a[i];
			if(now==0){
				now++;
				res++;
			}else if(now<0){
				ll b=-now+1LL;
				res+=b;
				now=1;
			}
		}else if(now>0){
			if(now+a[i]<0){
				now=now+a[i];
			}else{
				ll b=now+a[i]+1LL;
				now=-1;
				res+=b;
			}
		}else{
			if(now+a[i]>0){
				now=now+a[i];
			}else{
				ll b=-(now+a[i])+1LL;
				now=1;
				res+=b;
			}
		}
	}
	ll ans=res;
	res=0;
	now=0;
	for(int i=0;i<n;i++){
		if(now==0){
			now+=a[i];
			if(now==0){
				now--;
				res++;
			}else if(now>0){
				ll b=now+1LL;
				res+=b;
				now=-1;
			}
		}else if(now>0){
			if(now+a[i]<0){
				now=now+a[i];
			}else{
				ll b=now+a[i]+1LL;
				now=-1;
				res+=b;
			}
		}else{
			if(now+a[i]>0){
				now=now+a[i];
			}else{
				ll b=-(now+a[i])+1LL;
				now=1;
				res+=b;
			}
		}
	}
	ans=min(res,ans);
	printf("%lld\n",ans);
	return 0;
}