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>
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)
#define ALL(x) (x).begin(),(x).end() 
#define SIZE(x) ((ll)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define INF 1000000000
using namespace std;
typedef long long ll;

long Even(vector<int> a){
    long res = 0;
    long temp = 0;

    REP(i, a.size()){
        temp += a[i];
        if (i % 2 == 0){
            while (temp <= 0){
            temp += 1;
            res += 1;
            }
        }
        else{
            while (temp >= 0){
            temp += -1;
            res += 1;
            }
        }
    }
    return res;
}

long Odd(vector<int> a){
    long res = 0;
    long temp = 0;

    REP(i, a.size()){
        temp += a[i];
        if (i % 2 == 0){
            while (temp >= 0){
            temp += -1;
            res += 1;
            }
        }
        else{
            while (temp <= 0){
            temp += 1;
            res += 1;
            }
        }
    }
    return res;
}

int main() {
    int n, ans;
    cin >> n;
    vector<int> a(n);
    REP(i, n){
        cin >> a[i];
    }
    
    cout << min(Even(a), Odd(a)) << endl;
}