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 <cmath>

using namespace std;
typedef long long ll;
const int MAXN = 100010;

int n;
ll a[MAXN];
ll s[MAXN];

ll countNum(bool direction) {
  ll cnt = 0;
  ll carry = 0;
  if (a[0] == 0) {
    a[0] = 1;
    cnt = 1;
    carry = 1;
  }
  cnt = direction ? cnt : abs(a[0]);
  carry = direction ? carry : -a[0];
  s[0] = direction ? a[0] : -a[0];
  for (int i=1;i<n;i++) {
    s[i] = s[i-1] + a[i];
  }
  for (int i=1;i<n;i++) {
    if ((s[i] + carry) * s[i-1] >= 0) {
      cnt += abs(s[i]+carry) + 1;
      if (s[i-1] < 0) {
        carry += abs(s[i]+carry) + 1;
        s[i] = 1;
      } else {
        carry -= abs(s[i]+carry) + 1;
        s[i] = -1;
      }
    } else {
      s[i] += carry;
    }
  }
  return cnt;
}

void solve() {
  ll maxCnt = min(countNum(true), countNum(false));
  cout << maxCnt << endl;
}

int main() {
  cin >> n;
  for (int i=0;i<n;i++) {
    cin >> a[i];
  }
  solve();
}