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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

program ec12;
var 
	a,s:array[0..100000] of longint;
	n,m,i,j,ans,sum1,sum2,ans1:longint;
begin 	
	readln(n);
	ans:=0;
	ans1:=0;
	s[0]:=0;
	for i:=1 to n do
	read(a[i]);
	if a[1]>0 then 
	begin 
		sum1:=a[1];
		ans:=0;
		ans1:=a[1]+1;
		sum2:=-1;
	end
	else
	begin
		if a[1]=0 then 
		begin 
			sum1:=1;
			sum2:=-1;
			ans:=1;
			ans1:=1;
		end
		else
		begin
			ans:=(-a[1])+1;
			sum1:=1;
			ans1:=0;
		end;
	end;
	for i:=2 to n do 
	begin 
		if sum1>0 then 
		begin 
			if sum1+a[i]>=0 then 
			begin 	
				inc(ans,sum1+a[i]+1);
				sum1:=-1;
			end
			else
			sum1:=sum1+a[i];
		end
		else
		begin 	
			if sum1+a[i]<=0 then 
			begin 
				inc(ans,abs(sum1+a[i])+1);
				sum1:=1;
			end
			else
			sum1:=sum1+a[i];
		end;
		if sum2>0 then 
		begin 
			if sum2+a[i]>=0 then 
			begin 	
				inc(ans1,sum2+a[i]+1);
				sum2:=-1;
			end
			else
			sum2:=sum2+a[i];
		end
		else
		begin 	
			if sum2+a[i]<=0 then 
			begin 
				inc(ans1,abs(sum2+a[i])+1);
				sum2:=1;
			end
			else
			sum2:=sum2+a[i];
		end;
	end;
	if ans<ans1 then 
	writeln(ans)
	else
	writeln(ans1);
end.