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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <utility>
#include <numeric>
#include <cmath>
using namespace std;
//変数デバッグ
#define DEB(variable) cout << #variable << '=' << variable << endl

//for簡易表記(引数ミス防止)
#define FOR(LoopVariable,numberOFbegin,numberOFend) for (int LoopVariable = (numberOFbegin); (LoopVariable) < (numberOFend); (LoopVariable)++)
#define REP(LoopVariable,numberOFend) for(int LoopVariable = 0;(LoopVariable)<(numberOFend);LoopVariable++)
int n;
vector<long long int> v;
long long int solve(int first_pon){
    long long int pon = first_pon, sum = 0, result = 0;
    REP(i,n){
        sum += v[i];
        if(pon>0&&sum>=0){
            result += sum + 1;
            sum = -1;

        }else if(pon<0&&sum<=0){
            result += (-sum) + 1;
            sum = 1;

        }
        if(sum>0){
            pon = 1;
        }else{
            pon = -1;
        }
    }
    return result;
}

int main(){

    
    cin >> n ;
    v = vector<long long int>(n);
    REP(i,n){
        cin >> v[i];
    }
    cout << min(solve(1), solve(-1)) << endl;
    return 0;
}