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 C (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{
char a[200000];
while(1==scanf("%s",a))
{
	int length,i,ok,j,sum=0;
	char temp;
	for(length=0;a[length]!='\0';length++);
	for(i=0;2*i<=length+1;i++)
	{
		ok=0;
		while(a[ok++]=='W');
		for(j=ok-1;j<length-1;j++)
		if(a[j]=='B'&&a[j+1]=='W')
		{
			temp=a[j];
			a[j]=a[j+1];
			a[j+1]=temp;
			sum++;
		}
	}
	printf("%d",sum);
}
return 0;
 }