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 in = new Scanner(System.in);
		String str = in.nextLine();
		int[] arr = new int[str.length()];
		int a=0;
		for(int i=0; i<str.length(); i++){
			if(str.charAt(i)=='B'){
				arr[i]=1;
			}
		}
		for(int j=0; j<str.length(); j++){
			for(int i=0; i<str.length()-1; i++){
				if(arr[i]==1&&arr[i+1]==0){
					arr[i]=0;arr[i+1]=1;
					a++;
				}
			}
		}		
		System.out.println(a);
	}

}