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.*;
public class Main {
	public static void main(String[] args)
	{
		//変数
		long count=0;
		boolean sw=false;
		String sb="";
		String sa="";
		String c="";
		
		//入力
		Scanner sc = new Scanner(System.in);
		String S=sc.next();
		
		//処理
		sb=S;
		while(sw==false)
		{
			c=sb.replace("BW","");
			sa=sb.replace("BW","WB");
			//System.out.println(sa);
			//System.out.println(c);
			//System.out.println("--------");
			if(sa.equals(sb))
			{
				sw=true;
			}else
			{
				sb=sa;
				count=count+(sa.length()-c.length())/2;
				//System.out.println(sb);
				//System.out.println(count);
			}
		}
		// 出力
		System.out.println(count);
	}
}