output_description: Print the maximum possible number of times the operation can be performed.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      S
attempt: import sys
  
  
  ## io ##
  def IS():
      return sys.stdin.readline().rstrip()
  
  
  def II():
      return int(IS())
  
  
  def MII():
      return list(map(int, IS().split()))
  
  
  def MIIZ():
      return list(map(lambda x: x - 1, MII()))
  
  
  ## dp ##
  def DD2(d1, d2, init=0):
      return [[init] * d2 for _ in range(d1)]
  
  
  def DD3(d1, d2, d3, init=0):
      return [DD2(d2, d3, init) for _ in range(d1)]
  
  
  ## math ##
  def to_bin(x: int) -> str:
      return format(x, "b")  # rev => int(res, 2)
  
  
  def to_oct(x: int) -> str:
      return format(x, "o")  # rev => int(res, 8)
  
  
  def to_hex(x: int) -> str:
      return format(x, "x")  # rev => int(res, 16)
  
  
  MOD = 10**9 + 7
  
  
  def divc(x, y) -> int:
      return -(-x // y)
  
  
  def divf(x, y) -> int:
      return x // y
  
  
  def gcd(x, y):
      while y:
          x, y = y, x % y
      return x
  
  
  def lcm(x, y):
      return x * y // gcd(x, y)
  
  
  def enumerate_divs(n):
      """Return a tuple list of divisor of n"""
      return [(i, n // i) for i in range(1, int(n**0.5) + 1) if n % i == 0]
  
  
  def get_primes(MAX_NUM=10**3):
      """Return a list of prime numbers n or less"""
      is_prime = [True] * (MAX_NUM + 1)
      is_prime[0] = is_prime[1] = False
      for i in range(2, int(MAX_NUM**0.5) + 1):
          if not is_prime[i]:
              continue
          for j in range(i * 2, MAX_NUM + 1, i):
              is_prime[j] = False
      return [i for i in range(MAX_NUM + 1) if is_prime[i]]
  
  
  ## libs ##
  from itertools import (
      accumulate as acc,
      combinations as combi,
      product,
      combinations_with_replacement as combi_dup,
  )
  from collections import deque, Counter
  from heapq import heapify, heappop, heappush
  from bisect import bisect_left
  
  # ======================================================#
  
  
  def main():
      s = IS()
      c = s.split("B")
      if s.index("W") != 0:
          c = c[1:]
      c = c[::-1]
      cnt_w = [len(ci) for ci in c]
      acc_w = acc(cnt_w)
      print(sum(acc_w))
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  There are N Reversi pieces arranged in a row. (A _Reversi piece_ is a disc
  with a black side and a white side.) The state of each piece is represented by
  a string S of length N. If S_i=`B`, the i-th piece from the left is showing
  black; If S_i=`W`, the i-th piece from the left is showing white.
  
  Consider performing the following operation:
  
    * Choose i (1 \leq i < N) such that the i-th piece from the left is showing black and the (i+1)-th piece from the left is showing white, then flip both of those pieces. That is, the i-th piece from the left is now showing white and the (i+1)-th piece from the left is now showing black.
  
  Find the maximum possible number of times this operation can be performed.
samples: [{"input": "BBW", "output": "2\n    \n\nThe operation can be performed twice, as follows:\n\n  * Flip the second and third pieces from the left.\n  * Flip the first and second pieces from the left.\n\n* * *"}, {"input": "BWBWBW", "output": "6"}]