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 {
    static Scanner sc = new Scanner(System.in);


    public static void main(String[] args) {
        char[] arr = sc.next().toCharArray();
        int count = 0;
        for (int i = 1; i < arr.length; i++) {
            for (int j = arr.length - 1; j >= i; j--) {
                if (arr[j - 1] == 'B' && arr[j] == 'W') {
                    char tmp = arr[j - 1];
                    arr[j - 1] = arr[j];
                    arr[j] = tmp;
                    count++;
                }
            }
        }
        System.out.println(count);
    }
}