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;

int main(){
    int n,chk;
    long long ans=0, ans2=0;
    scanf("%d",&n);
    vector<int> a(n);
    for (auto&e:a) scanf("%d",&e);
    chk=a[0];
    for (int i=1;i<n;i++) {
        if (i%2) {
            chk+=a[i];
            if (chk>=0) {
                ans+=chk+1;
                chk=-1;
            }
        } else {
            chk+=a[i];
            if (chk<=0) {
                ans+=-1*chk+1;
                chk=1;
            }
        }
    }
    chk=a[0];
    for (int i=1;i<n;i++) {
        if (i%2==0) {
            chk+=a[i];
            if (chk>=0) {
                ans2+=chk+1;
                chk=-1;
            }
        } else {
            chk+=a[i];
            if (chk<=0) {
                ans2+=-1*chk+1;
                chk=1;
            }
        }
    }
    if (a[0]==0) printf("%lld\n",min(ans,ans2));
    else if (a[0]>0) printf("%lld\n",ans);
    else printf("%lld\n",ans2);
    return 0;
}