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)

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


using namespace std;


using ll = long long;


int main()
{
    ll acc;
    ll N;
    vector<ll> A;
    cin >> N;
    for (ll i = 0; i < N; i++) {
        ll tmp;
        cin >> tmp;
        A.push_back(tmp);
    }

    ll ans_pos = 0;
    if (A[0] <= 0) {
        acc = 1;
        ans_pos += abs(A[0]) + 1;
    } else {
        acc = A[0];
    }
    for (ll i = 1; i < N; i++) {
        ll next = acc + A[i];
        if (acc * next >= 0) {
            ans_pos += abs(next) + 1;
            next = -1 * (acc / abs(acc));
        }
        // // cout << "acc: " << acc << ", next: " << next << endl;
        acc = next;
    }

    ll ans_neg = 0;
    acc = A[0];
    if (acc >= 0) {
        acc = -1;
        ans_neg += abs(A[0]) + 1;
    } else {
        acc = A[0];
    }
    for (ll i = 1; i < N; i++) {
        ll next = acc + A[i];
        if (acc * next >= 0) {
            ans_neg += abs(next) + 1;
            next = -1 * (acc / abs(acc));
        }
        // // cout << "acc: " << acc << ", next: " << next << endl;
        acc = next;
    }

    ll ans = min(ans_pos, ans_neg);
    cout << ans << endl;
}