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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

object Main extends App {
  private def func(s:Array[Char]):Int = {
    var count = 0
    var find = false
    do{
      var i=0
      find = false
      while(i < s.size-1 /*&& !find*/){
        if(s(i)=='W' && s(i+1)=='B'){
          s(i  )='B'
          s(i+1)='W'
          find=true
          count +=1 
        }else{
          i+=1
        }
        //println(count, i, find, s.toSeq)
      } 
    } while(find)    
    count
  }
  println(func(io.Source.stdin.getLines.toIndexedSeq.head.toCharArray.reverse))
}