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 ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
const ll INF = 1e9;
const ll MOD = 1e9 + 7;
int main()
{
    ll n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) { cin >> a[i]; }
    vector<ll> sum(n);
    sum[0] = a[0];
    for (ll i = 1; i < n; i++) {
        sum[i] += sum[i-1] + a[i];
    }
    ll even_sum = 0;
    ll odd_sum = 0;

    // even should be plus
    rep(i, n)
    {
        if (i % 2 == 0 && sum[i] >= 0) { // odd & plus
            even_sum += 1 + sum[i];
        }
        if (i % 2 == 1 && sum[i] <= 0) { // even & minus
            even_sum += 1 - sum[i];
        }
    }
    // odd should be plus
    rep(i, n)
    {
        if (i % 2 == 0 && sum[i] <= 0) { // odd & minus
            odd_sum += 1 - sum[i];
        }
        if (i % 2 == 1 && sum[i] >= 0) { // even & plus
            odd_sum += 1 + sum[i];
        }
    }
    cout << min(odd_sum, even_sum) << endl;
    return 0;
}