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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>

int main(){
  char S[200000];
  int N=0;
  long ans=0;
  
  while (scanf("%c",&S[N])!=EOF){N++;}
  N--;

  int L=0,R=N;
  while (1)
  {
    if (R>L)
    {
      while (S[L]!='B'&&L<N){L++;}
      while (S[R]!='W'&&R>0){R--;}
      if(R>L){
        ans += R-L;
        L++;
        R--;
      }
    }else{
      break;
    }
  }
  
  printf("%ld",ans);

  return 0;
}