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 OCaml (4.02.3) (judge verdict: Accepted)

let id x = x

let init n f =
  let rec g i ns = if i = 0 then 0 :: ns else g (i - 1) (i :: ns) in
  List.map f (g n [])

let split_string ?(pattern="") = Str.split @@ Str.regexp pattern

let ss = split_string @@ Scanf.scanf "%s\n" id

let len = List.length ss

let () =
  ss
    |> List.mapi (fun idx s -> (len - idx - 1, s))
    |> (fun lst -> List.fold_right (fun (idx, s) (sum, b_cnt) -> if s = "B" then (sum + idx - b_cnt, b_cnt + 1) else (sum, b_cnt)) lst (0, 0))
    |> (fun (sum, _) -> sum)
    |> print_int;
  print_newline ()