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 <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <array>
#include <vector>
#include <functional>
#include <numeric>
#include <limits>
using namespace std;

int main(void) {
	long long n, i, j, sw, sw2, count = 0, add = 0, adda = 0;
		cin >> n;
	vector<long long> a(n);
	for (i = 0; i < n; i++) {
		cin >> a[i];
		adda += a[i];
	}
	if (a[0] == 0) {
		a[0]++; count++; add++;
	}
	if (a[0] > 0)sw = 1;
	else sw = -1;
	add += a[0];



	if ((adda > 0 && n % 2 == 1) || (adda < 0 && n % 2 == 0)) {
	}
	else {
		if (a[0] < 0) {
			while (a[0] != 1) {
				add++; a[0]++; count++;
			}
		}
		else {
			while (a[0] != -1) {
				add--; a[0]--;count++;
			}
		}
	}
	if (a[0] > 0)sw = 1;
	else sw = -1;
	for (i = 1; i < n; i++) {
		add += a[i];
		
		if (sw == 1) {
			if (add < 0) {
			}
			else {
				while (add != -1) {
					a[i]--; add--; count++;
				}
			}
		}
		else {
			if (add > 0) {
			}
			else {
				while (add != 1) {
					a[i]++; add++; count++;
				}
			}
		}
		if (a[i] > 0)sw = 1;
		else sw = -1;
	}
	cout << count << endl;
	return 0;
}