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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii =  pair<int, int>;
using pll =  pair<long long, long long>;
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007LL;
constexpr long long INF = 1000000009LL;
constexpr long long LINF = 1000100010001000100LL;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define rept(i, j, n) for(int i=(j); i<(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }


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

    ll res = 1e18+10;
    //正,負,正,負,...
    ll sum = 0, cost = 0;
    rep(i, n){
        sum += A[i];
        if(i%2==0 && sum <= 0){
            ll num = -sum + 1;
            sum += num; cost +=num;
        }else if(i%2==1 && sum >= 0){
            ll num = sum + 1;
            sum -= num; cost += num;
        }
    }
    chmin(res, cost);
    //負,正,負,正,...
    sum = 0, cost = 0;
    rep(i, n){
        sum += A[i];
        if(i%2==1 && sum <= 0){
            ll num = -sum + 1;
            sum += num; cost +=num;
        }else if(i%2==0 && sum >= 0){
            ll num = sum + 1;
            sum -= num; cost += num;
        }
    }
    chmin(res, cost);

    cout << res << ln;
}