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){
        Scanner sc = new Scanner(System.in);
        String S = sc.next();
        int [] list = new int [S.length()];
        for(int a=0;a<S.length();a++){
            if(S.charAt(a)=='B') list[a]=0;
            else list[a]=1;
        }
        int num=0;
        for(;;){
            int check = 0;
            for(int a=0;a<S.length()-1;a++){
                if (list[a] == 0 && list[a + 1] == 1) {
                    list[a] = 1;
                    list[a + 1] = 0;
                    num++;
                    check=1;
                }
            }
            if(check==0)break;
        }
        System.out.println(num);
    }
}