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)

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
using namespace std;
typedef long long int llint;
typedef vector<int> vint;
typedef vector<llint> vllint;
typedef vector<pair<int, int>> vpint;
typedef vector<pair<llint, llint>> vpllint;
#define rep(i,n) for(int i=0;i<n;i++)
#define yes(ans) if(ans)cout<<"yes"<<endl;else cout<<"no"<<endl;
#define Yes(ans) if(ans)cout<<"Yes"<<endl;else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl;else cout<<"NO"<<endl;
#define POSSIBLE(ans) if(ans)cout<<"POSSIBLE"<<endl;else cout<<"IMPOSSIBLE"<<endl;

class UnionFind {
public:
	//親の番号を格納する。親だった場合は-(その集合のサイズ)
	vector<int> Parent;

	//作るときはParentの値を全て-1にする
	//こうすると全てバラバラになる
	UnionFind(int N) {
		Parent = vector<int>(N, -1);
	}

	//Aがどのグループに属しているか調べる
	int root(int A) {
		if (Parent[A] < 0) return A;
		return Parent[A] = root(Parent[A]);
	}

	//自分のいるグループの頂点数を調べる
	int size(int A) {
		return -Parent[root(A)];//親をとってきたい
	}

	//AとBをくっ付ける
	bool connect(int A, int B) {
		//AとBを直接つなぐのではなく、root(A)にroot(B)をくっつける
		A = root(A);
		B = root(B);
		if (A == B) {
			//すでにくっついてるからくっ付けない
			return false;
		}

		//大きい方(A)に小さいほう(B)をくっ付けたい
		//大小が逆だったらひっくり返しちゃう。
		if (size(A) < size(B)) swap(A, B);

		//Aのサイズを更新する
		Parent[A] += Parent[B];
		//Bの親をAに変更する
		Parent[B] = A;

		return true;
	}
};

//aとbの最大公約数を求めるよ
long long GCD(long long a, long long b) {
	if (b == 0) return a;
	else return GCD(b, a % b);
}

// 返り値: a と b の最大公約数
// ax + by = gcd(a, b) を満たす (x, y) が格納される
long long extGCD(long long a, long long b, long long &x, long long &y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	long long d = extGCD(b, a%b, y, x);
	y -= a / b * x;
	return d;
}

bool check(char a, char b) {
	return a > b;
}

int main() {
	int n;
	cin >> n;
	vllint a(n);
	rep(i, n) {
		cin >> a[i];
	}
	llint cnt = 0;
	llint now = 0;
	rep(i, n) {
		now += a[i];
		if (i % 2 == 0) {
			if (now <= 0) {
				cnt += 1 - now;
				now = 1;
			}
		}
		else {
			if (now >= 0) {
				cnt += now + 1;
				now = -1;
			}
		}
	}
	llint ans = cnt;
	cnt = 0;
	now = 0;
	rep(i, n) {
		now += a[i];
		if (i % 2 == 1) {
			if (now <= 0) {
				cnt += 1 - now;
				now = 1;
			}
		}
		else {
			if (now >= 0) {
				cnt += now + 1;
				now = -1;
			}
		}
	}
	ans = min(cnt, ans);
	cout << ans << endl;
	return 0;
}