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: Accepted)

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef long long int lli;

int main() {
  int i, j;
  // scanf("%[^\n]%*c", a);
  lli n;
  char str[200001];

  scanf("%[^\n]%*c", str);

  i = 0;
  lli ind_sum = 0;
  lli wc = 0;
  while (str[i] != '\0') {
    if (str[i] == 'W') {
      ind_sum += i;
      wc += 1;
    }
    i++;
  }

  lli mini = (wc - 1) * (wc) / 2;

  printf("%lld", ind_sum - mini);

  return 0;
}