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;
//template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
//template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

/* attention
    long longのシフト演算には気をつけよう
    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも
*/


int main(){
    int n; cin >> n;
    vector<ll> a(n);
    for(int i=0; i<n; i++) cin >> a[i];

    ll ans = 1000000007;
    ll sum = 0;
    ll tmp = 0;
    // - + - + ...
    for(int i=0; i<n; i++){
        sum += a[i];
        if(i%2 == 0 && sum >= 0){
            tmp += sum + 1;
            sum = -1;
        }
        else if(i%2 == 1 && sum <= 0){
            tmp += 1 - sum;
            sum = 1;
        }
    }

    ans = min(ans, tmp);

    tmp = 0;
    sum = 0;
    // + - + - + ...
    for(int i=0; i<n; i++){
        sum += a[i];
        if(i%2 == 0 && sum <= 0){
            tmp += 1 - sum;
            sum = 1;
        }
        else if(i%2 == 1 && sum >= 0){
            tmp += 1 + sum;
            sum = -1;
        }
    }

    ans = min(ans, tmp);

    cout << ans << endl;
}