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: Wrong Answer)

import java.math.BigInteger;
import java.util.*;

public class Main {
	
	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			char [] data = sc.next().toCharArray();
			int count = 0;
			int sum = 0;
			for(int i = 0; i < data.length; i ++){
				if(data[i] == 'B'){
					sum += Math.max(0, data.length - 1 - i - count);
					count++;
				}
			}
			System.out.println(sum);
		}
	}

	private void debug(Object... o) {
		System.out.println("debug = " + Arrays.deepToString(o));
	}

	public static void main(String[] args) {
		new Main().doit();
	}

}