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.*;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    String s = sc.next();
    char[] S = s.toCharArray();
    int tmp;
    boolean flag = (S[0] == 'B');
    tmp = 100;
    if(flag){
      tmp = 0;
    }
    long count = 0;
    for(int i = 1; i< S.length; i++){
      if(flag){
        if(S[i] == 'W'){
          S[tmp] = 'W';
          S[i] = 'B';
          count += i - tmp;
          tmp = tmp + 1;
        }
      }else{
        if(S[i] == 'B'){
          tmp = i;
        }
      }
      flag = (S[i] == 'B');
    }
    System.out.println(count);
  }
}