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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		String s = sc.next();
		
		StringBuilder sb = new StringBuilder(s);
		
		int count = 0;
		
		while(true) {
			boolean b = false;
			for(int i = 1; i < s.length(); i++) {
				if(s.charAt(i-1) == 'B' && s.charAt(i) == 'W') {
					sb.setCharAt(i-1, getRev(s.charAt(i-1)));
					sb.setCharAt(i, getRev(s.charAt(i)));
					
					s = sb.toString();
					
					count++;
					b = true;
					break;
				}
			}
			if(b == false) {
				break;
			}
		}
		
		System.out.println(count);
	}
	
	public static char getRev(char c) {
		switch(c) {
		case 'B':
			return 'W';
		case 'W':
			return 'B';
		}
		return c;
	}
}