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;

#define debug(x) cerr << x << " = "(x) << endl;
#define ll long long
#define rep(i,x,y) for(int i=x;i<y;i++)
#define rel(i,x,y) for(int i=x-1;i>=y;i--)
#define all(x) x.begin(),x.end()

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    ll n; cin >> n;
    int a[n];
    rep(i,0,n) cin >> a[i];
    int rui[n];
    ll ans = 0;
    rui[0] = a[0];
    // + -の順
    if(rui[0]<=0){
        ans += abs(rui[0])+1;
        rui[0] = 1;
    }
    bool plus = false;
    rep(i,1,n){
        rui[i] = rui[i-1] +a[i];
        if(plus && rui[i]<=0){
            ans += abs(rui[i])+1;
            rui[i] = 1;
        }
        if(!plus && rui[i]>=0){
            ans += abs(rui[i])+1;
            rui[i] = -1;
        }
        plus ^= 1;
    }
    // - + の時
    ll tans = 0;
    rui[0] = a[0];
    if(rui[0]>=0){
        tans += abs(rui[0])+1;
        rui[0] = -1;
    }
    plus = true;
    rep(i,1,n){
        rui[i] = rui[i-1] +a[i];
        if(plus && rui[i]<=0){
            tans += abs(rui[i])+1;
            rui[i] = 1;
        }
        if(!plus && rui[i]>=0){
            tans += abs(rui[i])+1;
            rui[i] = -1;
        }
        plus ^= 1;
    }
    cout << min(ans,tans) << endl;
}