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.util.*;
public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        String S = sc.next();
        char[] chars = S.toCharArray();
        int[] stones = new int[chars.length];
        for(int i=0; i<chars.length-1; i++) {
            if(chars[i] == 'B') {
              stones[i] = 0;
            } else {
              stones[i] = 1;
            }
        }
        int max = 0;
        while(true) {
            int currentMax = max;
            for(int i=0; i<chars.length-1; i++) {
                if(stones[i] == 0 && stones[i+1] == 1) {
                    max ++ ;
                    stones[i] = 1;
                    stones[i+1] = 0;
                }else if(stones[i] == 1 && stones[i+1] == 0) {
                    i++;
                }
            }
            if (currentMax == max) {
                break;
            }
        }
        System.out.println(max);
    }
}