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: Time Limit Exceeded)

def BW_reverse(a, length)
  (0..(length-@count/2)).reverse_each do |i|
    if a[i] == "B" && a[i+1] == "W"
      a[i], a[i+1] = a[i+1], a[i]
      @count = @count + 1
      return BW_reverse(a, length)
    end
  end
end

a = gets.chomp.split("").map(&:to_s)
length = a.length - 2
@count = 0

BW_reverse(a, length)
puts @count