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>
#include <time.h>

int main(){
	//clock_t start,end;
    //start = clock();

	char a[200000];
	int i,j,k=0;
	int c=0;
	
	scanf("%s",a);
	
	for(i=0;a[i]!='\0';i++);
	
	for(j=0;j<i;j++){
	
	if(a[j]=='B' && a[j+1]=='B' && a[j+2]=='W'){
		a[j]='W';
		a[j+1]='B';
		a[j+2]='B';
		c = c+2;
		k=1;
	}else if(a[j]=='B' && a[j+1]=='W'){
		if(a[j+2]=='B' && a[j+3]=='W'){
		a[j]='W';
		a[j+1]='W';
		a[j+2]='B';
		a[j+3]='B';
		c = c+3;
		k=1;
		}else{
		a[j]='W';
		a[j+1]='B';
		c = c+1;
		k=1;
		}
	}
	
	
	if(j==i-1&&k==1){
		j=-1;
		k=0;
	}
	}
	
	printf("%d\n",c);
	
	//end = clock();
  	//printf("%.2f秒かかりました\n",(double)(end-start)/CLOCKS_PER_SEC);
	
	return 0;
}