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: Time Limit Exceeded)

#include <bits/stdc++.h>

#define REP(i, n) for(int i = 0; i < n; ++i)
using namespace std;

int main()
{   
    int n;
    cin >> n;

    vector<long> a(n);
    REP(i, n)   cin >> a[i];

    vector<bool> sign = {true, false};
    vector<int> cnt = {0, 0};

    long total;

    REP(k, 2){
        vector<long> a_tmp = a;

        for (int i = 0; i < n; ++i) {
            total = accumulate(a_tmp.begin(), a_tmp.begin()+i, 0) + a_tmp[i];

            if (sign[k] && total >= 0) {
                a_tmp[i] -= (total + 1);
                cnt[k] += (total + 1);
                sign[k] = false;
            }
            
            else if (!sign[k] && total <= 0) {
                a_tmp[i] += (abs(total) + 1);
                cnt[k] += (abs(total) + 1);
                sign[k] = true;
            }

            else
                sign[k] = !sign[k];
        }
    }

    cout << min(cnt[0], cnt[1]) << endl;
}