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 PHP7 (7.0.15) (judge verdict: Time Limit Exceeded)

<?php /*atcoder.php*/

$at = new Atcoder();
$at->main();
unset($at);

class Atcoder{
	private function lineIn(){
		$buf = fgets(STDIN);
		return $this->killNewLine($buf);
	}
	private function killNewLine($buf = null){
		$buf = str_replace("\r\n", "", $buf);
		$buf = str_replace("\n", "", $buf);
		$buf = str_replace("\r", "", $buf);
		return $buf;
	}

	public function main(){
		$s = $this->lineIn();
	
		$ans = 0;
		$i = 0;
		$max = count(str_split($s));

		while(true){
			if($s[$i] == "B"){
				if($s[($i + 1)] == "W"){
					$s[$i] = "W";
					$s[($i + 1)] = "B";
					$i = 0;
					$ans++;
				}else{
					$i++;
				}
			}else{
				$i++;
			}
			
			if(($i + 1) < $max){
				continue;
			}else{
				break 1;
			}
		}

		echo $ans . "\n";

	}
}
?>