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 <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <sstream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>
#include <deque>

using namespace std;

#define mod 1000000007

int n;
int a[100001];

int solve()
{
	long long int sum = a[0];
	long long int ans = 0;
	for(int i = 1; i < n; i++){
		if(sum > 0 && sum + a[i] >= 0){
			ans += abs(-1 - sum - a[i]);
			sum += -1 - sum;
		} else if(sum < 0 && sum + a[i] <= 0){
			ans += abs(1 - sum - a[i]);
			sum += 1 - sum;
		} else {
			sum += a[i];
		}
	}
	return ans;
}

int main()
{
	cin >> n;
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	long long int tmp1 = 0, tmp2 = 0;
	int tmp = a[0];
	if(a[0] <= 0){
		a[0] = 1;
		tmp1 += 1 - a[0];
	}
	tmp1 += solve();
	a[0] = tmp;
	if(a[0] >= 0){
		a[0] = -1;
		tmp2 += a[0] + 1;
	}
	tmp2 += solve();
	cout << min(tmp1, tmp2) << endl;
}