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 Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
  val sc = Scanner(System.`in`)
  val s = sc.next()

  var count = 0
  val array = s.toCharArray()
  while (array.joinToString("").contains("BW")) {
    count += replace(array)
  }

  println(count)

}

private fun replace(s: CharArray): Int {
  val n = s.size
  var count = 0
  for (i in 0 until n - 1 ) {
    if (s[i] == 'B' && s[i+1] == 'W') {
      count++
      s[i] = 'W'
      s[i+1] = 'B'
    }
  }

  return count
}