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>
using namespace std;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)

int main(){
    ll n;
    cin >> n;
    vector<ll> A(n), sum(n);
    rep(i, 0, n) cin >> A[i];

    // sum[0] > 0
    ll cnt1 = 0;
    if(A[0] > 0){
        sum[0] = A[0];
    }else{
        sum[0] = 1;
        cnt1 += abs(1-A[0]);
    }
    rep(i, 1, n){
        if(sum[i-1] > 0){
            if(sum[i-1] + A[i] < 0){
                sum[i] = sum[i-1] + A[i];
            }else{
                sum[i] = -1;
                cnt1 += abs(sum[i-1] + A[i] - (-1));
            }
        }else{
            if(sum[i-1] + A[i] > 0){
                sum[i] = sum[i-1] + A[i];
            }else{
                sum[i] = 1;
                cnt1 += abs(1 - (sum[i-1] + A[i]));
            }

        }
    }

    // sum[0] < 0
    ll cnt2 = 0;
    if(A[0] < 0){
        sum[0] = A[0];
    }else{
        sum[0] = -1;
        cnt2 += abs(A[0]-(-1));
    }
    rep(i, 1, n){
        if(sum[i-1] > 0){
            if(sum[i-1] + A[i] < 0){
                sum[i] = sum[i-1] + A[i];
            }else{
                sum[i] = -1;
                cnt2 += abs(sum[i-1] + A[i] - (-1));
            }
        }else{
            if(sum[i-1] + A[i] > 0){
                sum[i] = sum[i-1] + A[i];
            }else{
                sum[i] = 1;
                cnt2 += abs(1 - (sum[i-1] + A[i]));
            }

        }
    }
    
    cout << min(cnt1, cnt2) << endl;
    return 0;
}