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;
typedef long long ll;
#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(int i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(int i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(int i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define uniq(x) (x).erase(unique((x).begin(), (x).end()), (x).end())

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    ll ans = 0, csum = 0, nsum = a[0];
    
    if (a[0] >= 0) {
        rep(i, n) {
            csum = nsum;
            nsum += ((i == (n - 1)) ? 0 : a[i + 1]);
            
            if (csum == 0) {
                if ((i % 2) == 1) {
                    ans--;
                    nsum--;
                }
                else {
                    ans++;
                    nsum++;
                }
            }
            else if (((i % 2) == 1) && (csum > 0)) {
                ll d = abs(csum) + 1;
                ans += d;
                nsum -= d;
            }
            else if (((i % 2) == 0) && (csum < 0)) {
                ll d = abs(csum) + 1;
                ans += d;
                nsum += d;
            }
        }
    }
    else {
        rep(i, n) {
            csum = nsum;
            nsum += ((i == (n - 1)) ? 0 : a[i + 1]);
            
            if (csum == 0) {
                if ((i % 2) == 1) {
                    ans++;
                    nsum++;
                }
                else {
                    ans--;
                    nsum--;
                }
            }
            else if (((i % 2) == 1) && (csum < 0)) {
                ll d = abs(csum) + 1;
                ans += d;
                nsum += d;
            }
            else if (((i % 2) == 0) && (csum > 0)) {
                ll d = abs(csum) + 1;
                ans += d;
                nsum -= d;
            }
        }
    }
    
    cout << ans << endl;
    return 0;
}