Problem:
Score : 300 points

Problem 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.

Constraints

1 \leq |S| \leq 2\times 10^5

S_i=B or W

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum possible number of times the operation can be performed.

Sample Input 1

BBW

Sample Output 1

2

The operation can be performed twice, as follows:

Flip the second and third pieces from the left.

Flip the first and second pieces from the left.

Sample Input 2

BWBWBW

Sample Output 2

6

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

s = STDIN.gets.chomp

if !s.include?('B')
  puts 0
  exit 0
end

if !s.include?('W')
  puts 0
  exit 0
end

last_index = s.size - 1
sorted = last_index - s.rindex('W')

black_indices = []
s.chars.each_with_index do |char, i|
  if char == 'B'
    black_indices << i
  end
end

if sorted == black_indices.size
  puts 0
  exit 0
end

black_indices = black_indices[0..(last_index - sorted - 1)]

# p last_index
# p sorted
# p black_indices

sum = 0
black_indices.each do |i|
  sum += last_index - sorted - i
  sorted += 1
end

p sum