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 <bits/stdc++.h>
//---------------------------
using namespace std;
//---------------------------
#define REP(i,n) for(int i = 0; i < (n); i++)
#define P(x) cout << (x) << "\n"

#define MOD 1000000007 // 1e9+7
#define PI acos(-1.0)
#define ll long long int // 10^18
#define INF 1000000001 // 1e9+1

int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
//---------------------------

int main(){
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);

  // ifstream in("input.txt");
  // cin.rdbuf(in.rdbuf());

  int n;cin>>n;
  // int a[100000];
  int b[100001];
  REP(i,n){
    int l;
    cin>>l;
    // a[i]=l;
    b[i]=l;
  }
  REP(i,n)b[i+1]+=b[i];
  // REP(i,n)P(b[i]);
  // P("-----");
  ll sm=0;
  for(int i=1;i<n;i++){
    // P(b[i]);
    if(b[i-1]*b[i]<0)continue;
    // P(i);
    int target=(b[i-1]>0)?-1:1;
    sm+=abs(b[i]-target);
    int dif=-b[i]+target;
    for(int j=i;j<n;j++){
      b[j]+=dif;
      // P(b[j]);
    }
  }
  // P("-----");
  P(sm);
  return 0;
}