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 <bits/stdc++.h>

using namespace std;

using ll = long long;

int main() {
  int n;
  cin >> n;
  vector<ll> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i) a[i] += a[i - 1];
  }
  ll ans = 0, base = 0, ans1 = 0, base1 = 0;
  for (int i = 0; i < n; ++i) {
    if (i & 1) {
      if ((a[i] + base) * -1 <= 0) {
        ans += a[i] + base + 1;
        base -= a[i] + base + 1;
      }
    } else {
      if ((a[i] + base) * -1 >= 0) {
        ans += -a[i] - base + 1;
        base += -a[i] - base + 1;
      }
    }

    if (!(i & 1)) {
      if ((a[i] + base1) * -1 <= 0) {
        ans1 += a[i] + base1 + 1;
        base1 -= a[i] + base1 + 1;
      }
    } else {
      if ((a[i] + base1) * -1 >= 0) {
        ans1 += -a[i] - base1 + 1;
        base1 += -a[i] - base1 + 1;
      }
    }
  }
  cout << min(ans, ans1) << endl;
}