Problem:
Score : 800 points

Problem Statement

Given is a string S, where each character is 0, 1, or ?.

Consider making a string S' by replacing each occurrence of ? with 0 or 1 (we can choose the character for each ? independently).
Let us define the unbalancedness of S' as follows:

(The unbalancedness of S') = \max \{ The absolute difference between the number of occurrences of 0 and 1 between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\}

Find the minimum possible unbalancedness of S'.

Constraints

1 \leq |S| \leq 10^6

Each character of S is 0, 1, or ?.

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum possible unbalancedness of S'.

Sample Input 1

0??

Sample Output 1

1

We can make S'= 010 and achieve the minimum possible unbalancedness of 1.

Sample Input 2

0??0

Sample Output 2

2

Sample Input 3

??00????0??0????0?0??00??1???11?1?1???1?11?111???1

Sample Output 3

4

Solution in Ruby (2.7.1) (judge verdict: Accepted)

S = gets.chomp
L = S.length

s = 0; as = [s]
S.each_char{|c| s += c == '1' ? 1 : -1; as << s}
@bs = Array.new(L + 1)
@bs[-1] = as[-1]
(0 ... L).reverse_each{|i| @bs[i] = [as[i], @bs[i + 1]].max}
z = @bs[0]

def f(m)
  s, ad, ret = 0, 0, 0
  S.each_char.with_index do |c, i|
    if c == '1'
      s += 1
    elsif c == '0'
      s -= 1
      ret = [ret, s].min
    elsif ad + @bs[i + 1] + 2 <= m
      s += 1
      ad += 2
    else
      s -= 1
      ret = [ret, s].min
    end
  end
  ret
end

puts [z - f(z), z + 1 - f(z + 1)].min