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;

#define times(n, i)      uptil(0, n, i)
#define rtimes(n, i)     downto((n) - 1, 0, i)
#define upto(f, t, i)    for(int _##i = (t), i = (f); i <= _##i; i++)
#define uptil(f, t, i)   for(int _##i = (t), i = (f); i <  _##i; i++)
#define downto(f, t, i)  for(int _##i = (t), i = (f); i >= _##i; i--)
#define downtil(f, t, i) for(int _##i = (t), i = (f); i >  _##i; i--)
typedef long double LD;
#define long long long

#if defined(EBUG) && !defined(ONLINE_JUDGE)
#define debug true
#define ln << endl
#else
#define debug false
#define ln << '\n'
#endif
#define tb << '\t'
#define sp << ' '



signed main() { // long: 64bit
    if(!debug) {
        cin.tie(0);
        ios::sync_with_stdio(0);
    }

    int N;
    scanf("%d",&N);

    vector<int> A(N);
    times(N, i) scanf("%d",&A[i]);

    long ans = LLONG_MAX;

    times(2, k) {
        long a = 0, s = 0;
        times(N, i) {
            s += A[i];
            if(k == i%2 && s<=0) {
                a += 1 - s;
                s = 1;
            } else if(k != i%2 && s>=0) {
                a += s - (-1);
                s = -1;
            }
        }
        ans = min(ans, a);
    }

    cout << ans ln;

    return 0;
}