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;
#define FOR(i, a, b) for (int i = (a), _##i = (b); i < _##i; ++i)
#define REP(i, a) for (int i = 0, _##i = (a); i < _##i; ++i)
#define pb push_back
typedef long long LL;

int n;
LL a[100004];

LL solve(LL sum){
	
	LL res = 0;	
	FOR(i,2,n){	
		if(sum < 0){
			if( sum + a[i] > 0){			
				sum += a[i];
				continue;		
			}		
			res+= 1 - (sum + a[i]);
			sum = 1;
		}else{
			if( sum + a[i] < 0) {			
				sum += a[i];
				continue;		
			}
			res+= (sum + a[i]) + 1;
			sum = -1;
		}
	}	
	return res;
}

int main(){
//	freopen("test.in","r",stdin);
	cin>> n;
	REP(i,n){
		cin >> a[i];
	}
	LL sum = a[0] + a[1];
	if(sum){
		if( sum > 0){
			LL res = min( solve(-1) + sum + 1, solve(sum) );
			cout << min( solve(1) + sum - 1, res ) << endl;
		}else{
			LL res = min( solve(1)  - sum + 1, solve(sum) );
			cout << min( solve(-1)  - sum - 1, res ) << endl;
		}
	}else{
		cout << min( solve(1), solve(-1) ) + 1 << endl;
	}	
	return 0;
}