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: Runtime Error)

#include <stdio.h>

int main(void)
{
    char *S;
    int time_new=0;
    int time_old=0;
    int i,j;
    int N=0;
    char tmp;
    
    scanf("%s",S);
    for(j=0; S[j]!='\0'; j++){
        N++;
    }
    
    do{
        time_old=time_new;
        for(i=0; i<N-1; i++){
            if(S[i]=='B' && S[i+1]=='W'){
                S[i]='W';
                S[i+1]='B';
                time_new++;
                }
            }
        
       
    }while(time_new!=time_old);

    printf("%d\n",time_new);
    
    return 0;

}