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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<bitset>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<iomanip>
using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define INF 2e9
#define MOD 1000000007
#define LINF (long long)4e18
#define PI 3.1415926535

using ll = long long;
using Pi = pair<int,int>;
using Pl = pair<ll,ll>;



int main(){
  int N; cin >> N;
  vector<ll> a(N);
  rep(i,N) cin >> a[i];

  ll ans1 = 0;
  ll ans2 = 0;
  ll sum1 = 0;
  ll sum2 = 0;
  rep(i,N){ // 最初が正の場合
    if(i % 2 == 0){
      if(sum1 + a[i] <= 0){
        ans1 += abs(sum1 + a[i] - 1);
        sum1 = 1;
      }
      else sum1 += a[i];
    }
    if(i % 2 == 1){
      if(sum1 + a[i] >= 0){
        ans1 += abs(sum1 + a[i] + 1);
        sum1 = -1;
      }
      else sum1 += a[i];
    }
  }
  rep(i,N){ // 最初が負の場合
    if(i % 2 == 1){
      if(sum2 + a[i] <= 0){
        ans2 += abs(sum2 + a[i] - 1);
        sum2 = 1;
      }
      else sum2 += a[i];
    }
    if(i % 2 == 0){
      if(sum2 + a[i] >= 0){
        ans2 += abs(sum2 + a[i] + 1);
        sum2 = -1;
      }
      else sum2 += a[i];
    }
  }
  ll ans = min(ans1,ans2);
  cout << ans << endl;
}