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 <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define all(cont) cont.begin(), cont.end()
#define foreach(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define fore(i,a,b) for(int i=a,almo5t=b;i<almo5t;++i)
#define SZ(x) ((int)x.size())
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define FIN std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int N = 0;

typedef long long int ll;

using namespace std;
int opinions[15][15];

int main(){
    int n;cin>>n;
    ll values[n] = {0};
    ll sums[n+1] = {0};
    ll values2[n] = {0};
    ll sums2[n+1] = {0};
    sums[n+1] = 0;
    fore(i,0,n){
        ll val;cin>>val;
        values[i] = val;
        sums[i+1] = val + sums[i];
    }
    fore(i,0,n){
        values2[i] = values[i];
        sums2[i+1] = sums[i+1];
    }
    ll ans = 0;
    ll ans2 = 0;

    fore(i,0,n){
        sums[i+1] = sums[i]+values[i];
        //cout<<sums[i+1]<<"\n";
        if(i%2 == 0){
            if(sums[i+1]<=0){
                ans += abs(values[i]+sums[i])+1;
                values[i] = 1;
                sums[i+1] = 1;
            }
        }else{
            if(sums[i+1]>=0){
                ans += abs(values[i]+sums[i])+1;
                values[i] = -1;
                sums[i+1] = -1;
            }
        }
    }
    //cout<<ans<<" ";
    fore(i,0,n){
        sums2[i+1] = sums2[i]+values2[i];
        //cout<<sums[i+1]<<"\n";
        if(i%2 == 0){
            if(sums2[i+1]>=0){
                ans2 += abs(values2[i]+sums2[i])+1;
                values2[i] = -1;
                sums2[i+1] = -1;
            }
        }else{
            if(sums2[i+1]<=0){
                ans2 += abs(values2[i]+sums2[i])+1;
                values2[i] = 1;
                sums2[i+1] = 1;
            }
        }
    }
    //cout<<ans2<<" ";
    ans = min(ans,ans2);
    cout<<ans;
    
	return 0;
}