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)

//inlclude before define 
#define _USE_MATH_DEFINES

//include
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<cmath>
#include<type_traits>
#include<numeric>
#include<assert.h>
#include<deque>
#include<iomanip>
#include<set>
#include<tuple>
#include<map>

using namespace std;

//typedef
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<char> VC;
typedef pair<int, int> P1;
typedef long long int ll;

//define
#define INF 1e9
#define NUM 1000000007
#define all(x) begin(x),end(x)
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,a,b) for(int i=int(a);i<int(b);i++)
#define chmin(a,b) a=min((a),(b))
#define chmax(a,b) a=max((a),(b))

//function

/*ll gcd(ll a, ll b) {
	if (b == 0) return a;
	else return gcd(b, a%b);
}*/

/*ll lcm(ll a,ll b) {
	ll g = gcd(a, b);
	return a * (b / g);
}*/

/*int dig(int a) {
	int sum = 0;
	while (a) {
		sum += a % 10;
		a /= 10;
	}
	return sum;
}*/

int main () {
	int n;
	cin >> n;
	VI a (n);
	int ju = 0;
	rep (i, n) {
		cin >> a[i];
		if (i % 2 == 0 && a[i] > 0) ju++;
		if (i % 2 == 1 && a[i] < 0) ju++;
	}

	ll sum=0,ans=0;
	if (ju > n/2) {
		//cout << "a"<< endl;
		rep (i, n) {
			sum += a[i];
			//cout << sum << endl;
			if (sum <= 0 && i%2==0) {
				ans += abs( 1 - sum );
				sum = 1;
			} else if(sum>=0 && i%2==1){
				ans += abs (-1 - sum);
				sum = -1;
			}
			//cout << sum << " " << ans << endl;
		}
	} else {
		//cout << "b" << endl;
		rep (i, n) {
			sum += a[i];
			//cout << sum << " ";
			if (sum >= 0 && i % 2 == 0) {
				ans += abs (-1 - sum);
				sum = -1;
			} else if (sum <= 0 && i % 2 == 1) {
				ans += abs (1 - sum);
				sum = 1;
			}
			//cout << sum <<" "<<ans<< endl;
		}
	}
	cout << ans << endl;
	

}