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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cassert>
#include <cstdint>
#include <cctype>
#include <numeric>
#include <bitset>
#include <functional>

using namespace std;

using ll =  long long;
using Pll = pair<ll, ll>;
using Pii = pair<int, int>;

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr ll MOD = 1000000007;
constexpr long double EPS = 1e-10;
constexpr int dyx[4][2] = {
    { 0, 1}, {-1, 0}, {0,-1}, {1, 0}
};

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n;
    cin >> n;
    vector<ll> a(n+2, 0);
    for(int i=1;i<=n;++i) {
        cin >> a[i];
    }
    vector<ll> b(a);

    // + から
    ll ans1 = 0LL;
    // - から
    ll ans2 = 0LL;
    for(int i=1;i<=n;++i) {
        if(i % 2) {
            if(b[i] >= 0) {
                ans2 += b[i] + 1;
                b[i] = -1;
            }
            if(a[i] <= 0) {
                ans1 += -a[i] + 1;
                a[i] = 1;
            }
        } else {
            if(b[i] <= 0) {
                ans2 += -b[i] + 1;
                b[i] = 1;
            }
            if(a[i] >= 0) {
                ans1 += a[i] + 1;
                a[i] = -1;
            }
        }
        a[i+1] += a[i];
        b[i+1] += b[i];
    }

    cout << min(ans1, ans2) << endl;
}