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>
 
int main(void){
  int min = 0;
  int inter = 0;
  int max = 0;
  int i,j,q,r;
  int n[30];
  
  if( scanf("%d", &r) == 1);
  
  for(i = 0; i < r; i++){
    scanf("%d", &n[i]);
  }
  
  for(j = 0; j < r; j++){
    for(q = 1; q < r; q++){
    	if(n[j] < n[q]){
          max = n[q];
          if(min < n[j]){
            inter = n[j];
          }else{
            min = n[j];
          }
        }else{
          max = n[j];
          if(min < n[q]){
            inter = n[q];
          }else{
            min = n[q];
          }
        }
    }
  }
 return (0);
}