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;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < n; i++)

int main(void){
    ll n;
    cin >> n;

    ll a[n];
    ll sum[n];
    rep(i, n){
        cin >> a[i];
    }

    ll count = 0;
    rep(i, n){
        if(i == 0){
            sum[i] = a[i];
        }else{
            sum[i] = sum[i-1] + a[i];
            if(sum[i] != 0 && sum[i-1] == abs(sum[i-1]) && sum[i] == abs(sum[i])){
                count += sum[i] + 1;
                sum[i] = -1;
            }else if(sum[i] != 0 && sum[i-1] != abs(sum[i-1]) && sum[i] != abs(sum[i])){
                count += 1 - sum[i];
                sum[i] = 1;
            }
        }
        if(sum[i] == 0){
            count++;
            if(i == 0){
                if(a[i+1] == abs(a[i+1]))
                    sum[i] = -1;
                else
                    sum[i] = 1;
            }else{
                if(sum[i-1] == abs(sum[i-1]))
                    sum[i] = -1;
                else
                    sum[i] = 1;
            }
        }
    }

    cout << count << endl;
    return 0;
}