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: Accepted)

import java.util.*;

// Do not use int, use long!!

public class Main {
	public static void main(String args[]){
			Scanner sc = new Scanner(System.in);
			
			String s = sc.next();
			//1 operation , w mov left
			//so count each "w" how many "b" exists
			long countb = 0;
			long ans = 0;
			for (int i = 0; i < s.length(); i++) {
				if (s.charAt(i) == 'B') {
					countb++;
				}else{
					ans += countb;
				}
			}

			System.out.println(ans);
			
			sc.close();
		}
	
}