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 i,j,k,sum=0;
	char temp;
	for(i=0;a[i]!='\0';i++);
	for(j=1;j<i;j++)
	for(k=0;k<i-1;k++)
	if(a[k]=='B'&&a[k+1]=='W')
	{
		temp=a[k];
		a[k]=a[k+1];
		a[k+1]=temp;
		sum++;
	}
	printf("%d",sum);
}
return 0;
 }