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 <string>
#include <unordered_map>
#include <math.h>
#include <algorithm>
#include <deque>
#include <queue>
#include <climits>
using namespace std;
#define rep(i,l,r) for(int i=l;i<r;i++)
#define repb(i,r,l) for(int i=r;i>l;i--)
#define ll long long
#define ull unsigned long long
#define ve vector
#define umap unordered_map
#define iter iterator
const int MxN = 1e5;
ll ar[MxN];
int n;
ll solve() {    
    ll res = 0;
    ll sum = 0;
    for(int i=0;i<n;i++) {
        if (i % 2 == 0) {
            if (sum + ar[i] <= 0) {
                res += llabs(sum + ar[i] - 1);
                sum = 1;
            } else 
                sum += ar[i];
        } else {
            if (sum + ar[i] >= 0) {
                res += llabs(sum + ar[i] + 1);
                sum = -1;
            } else 
                sum += ar[i];
        }
    }
    return res;
}

int main() {
    cin>>n;
    for(int i=0;i<n;i++) cin>>ar[i];
    ll a =solve();
    for(int i=0;i<n;i++) ar[i] = -ar[i];
    ll b =solve();
    cout<<min(a, b)<<endl;
    return 0;
}