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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;

const ll INFTY = 1001001001;

int main()
{
    int N;
    cin >> N;
    vector<ll> a(N + 1), b(N + 1);
    for (int i = 0; i < N; ++i)
    {
        cin >> a[i];
        b[i] = a[i];
        if (i != 0)
        {
            a[i] += a[i - 1];
            b[i] += b[i - 1];
        }
    }

    ll ans = INFTY;
    for (int sign = 0; sign < 2; ++sign)
    {
        if (sign)
        {
            for (int i = 0; i < N + 1; ++i)
            {
                a[i] = b[i];
            }
        }
        ll v = 0, count = 0;
        for (int i = 0; i < N; ++i)
        {
            if (sign)
            {
                if (a[i] >= 0)
                {
                    count += abs(-a[i] - 1);
                    v -= abs(-a[i] - 1);
                }
                a[i + 1] += v;
            }
            else
            {
                if (a[i] <= 0)
                {
                    count += abs(-a[i] + 1);
                    v += abs(-a[i] + 1);
                }
                a[i + 1] += v;
            }
            sign = !(sign);
        }
        ans = min(ans, count);
    }
    cout << ans << endl;
}