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<algorithm>
#include<string>
#include<set>
#include<map>
#include<queue>
#define REP(i,a) for (int i = 0;i < (a);++i)
#define FOR(i,a,b) for (int i = (a);i < (b); ++i)
#define FORR(i,a,b) for (int i = (a);i >= (b); --i)
#define ALL(obj) (obj).begin,(obj).end()
#define SORT(list) sort(ALL((list)));
#define MOD 1000000007
using namespace std;
using ll = long long;
int main(){
    int n;
    cin >> n;
    ll a[n];
    REP(i,n)cin >> a[i];
    ll sum=0;
    ll answer = 0;
    REP(i,n){
        if(i%2 == 0){
            if(a[i] + sum <= 0){
                answer += llabs(sum+a[i])+1;
                sum = 1;
            }
            else{
                sum += a[i];
            }
        }
        else{
            if(a[i] + sum >= 0){
                answer += llabs(sum + a[i]) + 1;
                sum = -1;
            }
            else{
                sum += a[i];
            }
        }
    }

    ll answer2 = answer;
    answer = 0;
    sum = 0;
    REP(i,n){
        if(i%2 == 1){
            if(a[i] + sum <= 0){
                answer += llabs(sum+a[i])+1;
                sum = 1;
            }
            else{
                sum += a[i];
            }
        }
        else{
            if(a[i] + sum >= 0){
                answer += llabs(sum + a[i]) + 1;
                sum = -1;
            }
            else{
                sum += a[i];
            }
        }
    }
    answer = min(answer,answer2);
    cout << answer << endl;
    return 0;
}