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() {
  int i, N=0, length=0;
  char S[200001];
  
  scanf("%s", S);
  
  for(i = 0; S[i] != '\0'; ++i){
    ++length;
  }
  
  for(i = length - 1; i >= 0; --i){
    if(S[i] == 'B' && S[i+1] == 'W'){
      S[i] = 'W';
      S[i+1] = 'B';
      ++N;
      i = length - 1;
    }
  }
  
  printf("%d\n", N);
  
  return 0;
}