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 OCaml (4.02.3) (judge verdict: Wrong Answer)

(* O(n) *)
let n = Scanf.scanf " %d" @@ (+) 0
let a_s = Array.init n @@ fun _ -> Scanf.scanf " %d" @@ (+) 0
let ans = ref max_int
let _ =
  for m = 0 to 1 do
    let sum = ref a_s.(0) in
    let c = ref 0 in
    if a_s.(0) > 0 then if m = 0 then (sum := -1; c := abs !sum + 1) else ()
    else if m = 1 then (sum := 1; c := abs !sum + 1);
    for i = 1 to n - 1 do
      if i mod 2 = m then
        (sum := !sum + a_s.(i);
        if !sum >= 0 then (c := !c + abs !sum + 1; sum := -1))
      else
        (sum := !sum + a_s.(i);
        if !sum <= 0 then (c := !c + abs !sum + 1; sum := 1))
    done;
    ans := min !ans !c
  done;
  Printf.printf "%d\n" !ans