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: Wrong Answer)

#include<iostream>
#include<iomanip>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<map>
#include<numeric>
#include<set>
#include<string>
#include<utility>
#include<vector>

#define rep(i, a, n)  for(int i=a; i<n; i++)
#define per(i, a, n)  for(int i=n-1; i>=a; i--)
#define fill0(n)  setfill('0') << right << setw(n)
#define all(x)  (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define MAX 1000000000
using namespace std;
typedef long long ll;

ll n;
ll data[100001];

ll solve(){
  ll res = 0;
  ll part = 0;
  part+=data[0];
  bool isP = part>0?true:false;
  rep(i, 1, n){
    part+=data[i];
    if(isP){
      if(part>0){
        res+=part+1;
        part=-1;
      }
      else if(part==0){
        res++;
        part=-1;
      }
      isP=false;
    }
    else{
      if(part<0){
        res+=1-part;
        part=1;
      }
      else if(part==0){
        res++;
        part=1;
      }
      isP=true;
    }
  }
  return res;
}

int main(){
  cin >> n;
  ll ans = 0;
  rep(i, 0, n)  cin >> data[i];
  ans = solve();
  rep(i, 0, n)  data[i]*=-1;
  cout << min(ans, solve()) << endl;
  return 0;
}