output_description: Print the maximum possible final value of x in a sequence of operations such
  that S equals T in the end.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      T
attempt: s = input()
  
  
  def get(s):
      n = len(s)
      ans = 0
      a = 0
      for i in range(0, n, 2):
          if s[i] == "1":
              a += 1
      b = 0
      c = 0
      no = n
      for i in reversed(range(n)):
          if s[i] == "1":
              if i % 2 == 0:
                  a -= 1
                  b += 1
              else:
                  c += 1
          else:
              no -= 1
              ans += b + a
              b, c = c, b
      for i in range(no + 1):
          ans += (i + 1) // 2
      return ans
  
  
  ans = 0
  if "0" in s:
      i = s.index("0")
      t = s[:i] + s[i + 1 :]
      for i in range(0, len(s), 2):
          if s[i] == "1":
              ans += 1
      ans += get(t)
  ans = max(ans, get(s))
  print(ans)
  
problem_description: Statement
  
  Takahashi has an empty string S and a variable x whose initial value is 0.
  
  Also, we have a string T consisting of `0` and `1`.
  
  Now, Takahashi will do the operation with the following two steps |T| times.
  
    * Insert a `0` or a `1` at any position of S of his choice.
    * Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
  
  Print the maximum possible final value of x in a sequence of operations such
  that S equals T in the end.
samples: [{"input": "1101", "output": "5\n    \n\nBelow is one sequence of operations that maximizes the final value of x to 5.\n\n  * Insert a `1` at the beginning of S. S becomes `1`, and x is incremented by 1.\n  * Insert a `0` to the immediate right of the first character of S. S becomes `10`, and x is incremented by 1.\n  * Insert a `1` to the immediate right of the second character of S. S becomes `101`, and x is incremented by 2.\n  * Insert a `1` at the beginning of S. S becomes `1101`, and x is incremented by 1.\n\n* * *"}, {"input": "0111101101", "output": "26"}]