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)

#define _GLIBCXX_DEBUG
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>
#include <map>
using namespace std;
typedef long long ll;
#define ALL(v) (v).begin(), (v).end()
const int inf = 1e9;
const int mod = 1e9+7;

int n;
ll a[100010];

ll solve () {
	ll now_sum = 0;;
	ll res = 0;

	for (int i = 0; i < n; i++) {
		if (i % 2 == 0) {
			if (now_sum + a[i] <= 0) {
				res += abs(now_sum + a[i] - 1);
				now_sum = 1;
			} else {
				now_sum += a[i];
			}
		} else {
			if (now_sum + a[i] >= 0) {
				res += abs(now_sum + a[i] - (-1));
				now_sum = -1;
			} else {
				now_sum += a[i];
			}
		}
	}

	return res;
}

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

	ll ans = solve();

	for (int i = 0; i < n; i++) a[i] *= -1;
	ans = min(ans, solve());

	cout << ans << endl;
    return 0;
}