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: Accepted)

uses math ;
var a:array[1..100000] of int64 ;
      i,n:longint ;
			ans1,ans2,sum1,sum2:int64 ;
begin 
   readln(n) ;
	 for i:=1 to n do read(a[i]) ;
	 if a[1]>0 then 
	 begin 
	   sum1:=a[1];
		 ans2:=a[1]+1 ;
		 sum2:=-1 ;
	 end
	 else if (a[1]=0) then 
	 begin 
	   sum1:=1;
		 ans1:=1;ans2:=1 ;
		 sum2:=-1 ;
	 end
	 else 
	 begin 
	   sum1:=1;
		 ans1:=abs(a[1])+1 ;
		 sum2:=a[1] ;
	 end;
	 for i:=2 to n do 
	 begin 
	   if sum1>0 then
		 begin 
		   if a[i]+sum1>=0 then 
			 begin 
			   ans1:=ans1+a[i]+sum1+1 ;
				 sum1:=-1 ;
			 end
			 else sum1:=sum1+a[i] ;
		 end
		 else 
		 begin 
		   if a[i]+sum1<=0 then 
			 begin 
			   ans1:=ans1+abs(sum1+a[i])+1 ;
				 sum1:=1 ;
			 end
			 else sum1:=sum1+a[i] ;
		 end;
	 end;
	 if sum1=0 then inc(ans1) ;
	 for i:=2 to n do //这段跟上段一模一样
	 begin 
	   if sum2>0 then
		 begin 
		   if a[i]+sum2>=0 then 
			 begin 
			   ans2:=ans2+a[i]+sum2+1 ;
				 sum2:=-1 ;
			 end
			 else sum2:=sum2+a[i] ;
		 end
		 else 
		 begin 
		   if a[i]+sum2<=0 then 
			 begin 
			   ans2:=ans2+abs(sum2+a[i])+1 ;
				 sum2:=1 ;
			 end
			 else sum2:=sum2+a[i] ;
		 end;
	 end;
	 if sum2=0 then inc(ans2) ;
	 writeln(min(ans1,ans2)) ;
end.