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)

/*------------------------------------
........Bismillahir Rahmanir Rahim....
..........created by Abdul Aziz.......
------------------------------------*/
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <unordered_map>
#include <queue>
#define   mod  998244353
#define   int  long long 
#define   ld   long double
#define   pb   push_back
#define   vi  vector<int> 
#define   dbg(x)  cerr << #x << " = " << x << '\n'
#define   sz(x)  (int)x.size()
#define   ff   first
#define   ss   second
#define   pii  pair<int,int>
using namespace std;

inline void solve(){
    int n; cin>>n; 
    vi a(n);
    for (int &i: a) cin>>i;
    int p=0,ne=0,s=0;
    for (int i=0;i<n;i++){
        s += a[i];
        if (i&1){
            if (s>=0){
                p += s+1;
                s = -1;
            }
        }
        else {
            if (s<=0){
                p += abs(s)+1;
                s=1;
            }
        }
    }
    s=0;
    for (int i=0;i<n;i++){
        s += a[i];
        if (i%2==0){
            if (s>=0){
                ne += s+1;
                s=-1;
            }
        }
        else {
            if (s<=0){
                ne += abs(s)+1;
                s=1;
            }
        }
    }
    cout << min(p,ne) << endl;
}

signed main()
{
    int n=1; // cin>>n;
    while (n--) solve();
    return 0;
}