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 <algorithm>
#include <cmath>
#include <cstdio>
#include <functional>
#include <numeric>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <fstream>
#include <bitset>
#include <time.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<ll> V;
typedef complex<double> Point;

#define PI acos(-1.0)
#define EPS 1e-10
const ll INF = 1e9 + 7;
const ll MOD = 1e9 + 7;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,N) for(int i=0;i<(N);i++)
#define ALL(s) (s).begin(),(s).end()
#define EQ(a,b) (abs((a)-(b))<EPS)
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
#define fi first
#define se second

int n;
int a[100100];
ll sum[100100];
ll table[100100];

int main() {
	cin >> n;
	bool f = 0;
	REP(i, n) {
		cin >> a[i];
		if (i == 0 && a[i] < 0)f = 1;
		if (f)a[i] = -a[i];
	}
	ll ans = INF;
	REP(j, 2) {
		ll bor = 0, num = 0;
		REP(i, n) {
			if (i == 0)sum[i] = a[i];
			else sum[i] = sum[i - 1] + a[i];
			if (j == 0) {
				if (i % 2 == 0)table[i] = 1;
				else table[i] = -1;
			}
			else {
				if (i % 2 == 0)table[i] = 1;
				else table[i] = -1;
			}
		}
		REP(i, n) {
			sum[i] += bor;
			if (sum[i] == 0) {
				sum[i] = table[i];
				num++;
				bor += table[i];
				continue;
			}
			if (table[i] * sum[i] < 0) {
				bor += table[i] - sum[i];
				num += abs(table[i] - sum[i]);
			}
		}
		ans = min(ans, num);
	}
	cout << ans << endl;
}