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<string>
#include<vector>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<numeric>
#include<iomanip>
#include<deque>
#include<tuple>
#include<queue>
#include<map>
#include <cstdint>
#include <boost/multiprecision/cpp_int.hpp>


#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define Endl endl
#define F first
#define S second

namespace mp = boost::multiprecision;

using cpp_int = mp::cpp_int;
using ll = long long;
using namespace std;
int operation(int n,ll cnt,vector<ll>sum,vector<ll> a) {
	for (int i = 1; i < n; i++) {
		sum[i] = sum[i - 1] + a[i];
		if (sum[i - 1] > 0) {
			if (sum[i] >= 0) {
				cnt += sum[i] + 1;
				sum[i] = -1;
			}
		}
		else if (sum[i - 1] < 0) {
			if (sum[i] <= 0) {
				cnt += abs(sum[i]) + 1;
				sum[i] = 1;
			}
		}
	}
	return cnt;
}
int main() {
	int n;
	cin >> n;
	vector<ll>sum(n),a(n),sum2(n);
	rep(i, n) {
		cin >> a[i];
	}
	ll count = 0;
	ll count2 = 0;
	sum[0] = a[0];
	if (sum[0]>0) {
		sum2[0] = -1;
		count2 += sum[0] + 1;
	}
	else if (sum[0] < 0) {
		sum2[0] = 1;
		count2 += abs(sum[0])+1;
	}
	else {
		sum[0] = 1;
		count++;
		sum2[0] = -1;
		count2++;
	}
	cout<<min(operation(n, count, sum, a) 
		,operation(n, count2, sum2, a))<< endl;
}