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 PHP (5.6.30) (judge verdict: Accepted)

<?php
//error_reporting(0);
$N = trim(fgets(STDIN));
//list($N,$K) = explode(" ",trim(fgets(STDIN)));
//$a = explode(" ",trim(fgets(STDIN)));
//$H = explode(" ",trim(fgets(STDIN)));

$a = str_split($N);
$c = array_count_values($a);
$stotal = 0;
$len = count($a);
for($i=0;$i<$len;$i++) {
  if ($a[$i]=='B') {
    $stotal += $i;
  }
}
$total = 0;
$len = $len -1;
for($i=0;$i<$c['B'];$i++) {
  $total += $len-$i;
}
printf("%d\n",$total-$stotal);