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 Ruby (2.3.3) (judge verdict: Accepted)


# Bを制約を満たす最小操作後の配列とし、
# sb[k][n]をB[0]>0 (k=1), B[0]<0 (k=1) である場合のB[n-1]までの累積和とする、
# この符号は+/-交互になる必要があるので、
# (sb[k][n-1] > 0 => sb[k][n] < 0) => (sb[k][n-1] + A[n-1] - op < 0)
# (sb[k][n-1] < 0 => sb[k][n] > 0) => (sb[k][n-1] + A[n-1] + op > 0)
# とする必要があり、k,nによる状態から最小操作数opがきまる
# dp[k][n]は上記opの累積和


N = gets.to_i
A = gets.split.map(&:to_i)

sb = 2.times.map{ Array.new(N + 1, 0) }
dp = 2.times.map{ Array.new(N + 1, 0) }
(1 .. N).each do |n|
  a = A[n - 1]
  2.times do |k|
    if (k + n).odd?
      if sb[k][n-1] + a > 0
        dp[k][n] = dp[k][n-1]
        sb[k][n] = sb[k][n-1] + a
      else
        dp[k][n] = dp[k][n-1] + (sb[k][n-1] + a).abs + 1
        sb[k][n] = 1
      end
    else
      if sb[k][n-1] + a < 0
        dp[k][n] = dp[k][n-1]
        sb[k][n] = sb[k][n-1] + a
      else
        dp[k][n] = dp[k][n-1] + (sb[k][n-1] + a).abs + 1
        sb[k][n] = -1
      end
    end
  end
end
puts [dp[0][N], dp[1][N]].min