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<bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
	cin.sync_with_stdio(false);
	int n;
	cin >> n;
	vector<ll>a(n);
	ll sum=0;
	int flag = true;
	for (int i = 0;i < n;i++) {
		cin >> a[i];
		if (sum*(sum + a[i]) > 0) {
			flag = false;
		}
		sum += a[i];
		if (sum == 0) {
			flag = false;
		}
	}
	if (flag) {
		cout << 0 << endl;
		return 0;
	}
	ll count1 = 0;
	ll count2 = 0;
	sum = a[0];
	sum += a[1];
	if (sum == 0) {
		count1 += 1;
	}else if (sum < 0) {
		count1 += -1 - sum;
	}
	else {
		count1 += sum + 1;
	}
	sum = -1;
	for (int i = 2;i < n;i++) {
		if (sum < 0 && sum + a[i] < 0) {
			count1 += abs(sum) - a[i] + 1;
			sum = 1;
		}
		else if (sum > 0 && sum + a[i] > 0) {
			count1 += abs(sum + a[i] + 1);
			sum = -1;
		}
		else {
			sum += a[i];
		}
		if (sum == 0) {
			count1 += 1;
		}
	}
	sum = a[0];
	sum += a[1];
	if (sum == 0) {
		count2 += 1;
	}
	else if (sum < 0) {
		count2 += 1 - sum;
	}
	else {
		count2 += sum - 1;
	}
	sum = 1;
	for (int i = 2;i < n;i++) {
		if (sum < 0 && sum + a[i] < 0) {
			count2 += abs(sum) - a[i] + 1;
			sum = 1;
		}
		else if (sum > 0 && sum + a[i] > 0) {
			count2 += abs(sum + a[i] + 1);
			sum = -1;
		}
		else {
			sum += a[i];
		}
		if (sum == 0) {
			count2 += 1;
		}
	}
	
	cout << min(count1,count2) << endl;
	return 0;
}