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.Scanner;
public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		String[] SS = s.split("");
		int[] a = new int[SS.length];
		for(int i = 0; i < SS.length; i++) {
			if(SS[i].equals("B")) {
				a[i] =1;
			}else {
				a[i] = -1;
			}
		}
		int ans1 = 0;
		int ans2 = 0;
		int c = 0;
		if(a[0] == 1) {
			for(int i = 1; i < SS.length; i++) {
				if(a[i] == -1) {
					ans1 += i-c;
					c++;
				}
			}
		}else {
			for(int i = 1; i < SS.length; i++) {
				if(a[i] == 1) {
					ans1 += i-c;
					c++;
				}
			}
		}
		if(a[SS.length-1] == 1) {
			c = SS.length-1;
			for(int i =SS.length-1; i >= 1; i--) {
				if(a[i] == -1) {
					ans2 += c-i;
					c--;
				}
			}
		}else {
			c = SS.length-1;
			for(int i =SS.length-1; i >= 1; i--) {
				if(a[i] == 1) {
					ans2 += c-i;
					c--;
				}
			}
		}

		System.out.println(Math.max(ans1,  ans2));
		sc.close();
	}
}