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: Wrong Answer)

#include <stdio.h>
#include <string.h>
int main(void){
	char s[200010];
  	scanf("%s", s);
  
  	int len = strlen(s),count[len],total=0;
 	int num=-1,lap=0;
  
  	for(int i=num+1;i<len;i++){
     	if(s[i]=='W'){
          	for(int j=num+1;j<i;j++){
				if(s[j]=='B'){
						count[lap]++;
                }
              num=i;
              lap++;
            }
        }
    }
	int sum=0;
  	for(int i=0;i<lap+1;i++){
      sum+=(lap-i)*count[i];
    }
  	
  	printf("%d", sum);
}