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: def getBinVal(s):
      v = 0
      ar = list(s)
      # print(ar)
      # print(len(s))
  
      for j in range(len(s)):
          if j % 2 == 0:
              v += int(ar[j])
      # print('getdata:' + s + ' >>> ' + str(v))
      return v
  
  
  T = input()
  n = len(T)
  
  aT1 = list(T)
  
  zeros = aT1.count("0")
  ones = aT1.count("1")
  
  s = ""
  x = 0
  
  # one
  if ones > 0:
      for i in range(ones):
          s = s + "1"
          x += getBinVal(s)
  
  # zero
  idx = 0
  
  if zeros > 0:
      for k in range(zeros):
          idx = T.find("0", idx)
          s = s[0:idx] + "0" + s[idx:]
          x += getBinVal(s)
          idx = idx + 1
  
  # x+=getBinVal('11111101')
  # x+=getBinVal('111101101')
  # x+=getBinVal('0111101101')
  
  print(x)
  
  
  # for i in range(n):
  # 	s=s + aT1[i]
  # 	x+=getBinVal(s)
  
  
  # aT2=list(T)
  # aT2.reverse()
  # s=''
  # x=0
  # print(aT2)
  # for i in range(n):
  # 	s=aT2[i] + s
  # 	x+=getBinVal(s)
  # print(s)
  # print(x)
  
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"}]