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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

int main(void)
{
	char S[200000];//1<=S<=2*10^5
	int i,count=0,f=0,N,b;
	
	scanf("%s",&S);
	
	for(i=0;S[i]!='\0';i++){
	N+=1;
	}
	
	for(i=0;i<N;i++){
		while(S[i]=='B'&&S[i+1]=='W'){
			S[i]='W';
			S[i+1]='B';
			count+=1;
			i-=1;
			b+=1;
		}
		i+=b;
		b=0;
	}
	
    printf("%d\n",count);
}