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;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i < (n); i++)
#define all(a) begin(a),end(a)
template<class T> istream& operator >> (istream& s, vector<T>& v)
{ for (T& x: v) { s >> x; } return s;}
template<class T> void cout_h(const vector<T>& c, int n=INT_MAX)
{ rep(i, min(n, (int)c.size())) cout << c[i] << (i==c.size()-1?"\n":" "); }
template<class T> inline bool chmin(T& a, T b) {return a>b?a=b,true:false;}
template<class T> inline bool chmax(T& a, T b) {return a<b?a=b,true:false;}
#define D1(a) cout<<#a<<":"<<a<<"\n"
#define D2(a,b) cout<<#a<<":"<<a<<","<<#b<<":"<<b<<endl 
#define D3(a,b,c) cout<<#a<<":"<<a<<","<<#b<<":"<<b<<","<<#c<<":"<<c <<endl
#define D4(a,b,c,d) cout<<#a<<":"<<a<<","<<#b<<":"<<b<<","<<#c<<":"<<c<<","<<#d<<":"<<d<<endl 
#define MOD 1000000007

struct _I { _I() {cin.tie(0);ios::sync_with_stdio(false);}} __I;

int main()
{
    int n; cin >> n;
    vi a(n);
    vll s(n);
    rep(i, n) {
        cin >> a[i];
        s[i] = a[i] + (i==0?0:s[i-1]);
    }

    ll ans[2] = {0, 0};
    ll change[2] = {0, 0};
    rep(i, n) {
        rep(j, 2) {
            ll a = ((i+j)%2==1) ? -1 : 1;
            ll sum = s[i]+change[j];
            if (sum == 0) {
                change[j] += a;
                ans[j] += 1;
            } else if ((sum < 0 && (i+j)%2==0) || (sum > 0 && (i+j)%2==1)){
                change[j] += a - sum;
                ans[j] += abs(a - sum);
            }
        }
    }
    
    cout << min(ans[0], ans[1]) << endl;
}