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: Wrong Answer)

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

#define BUF_SIZE 200010
unsigned long count = 0;
char s[BUF_SIZE];
char *r;
int i;
int j;

int main(void){
    fgets(s, BUF_SIZE, stdin);
    for(i=0; i<BUF_SIZE; i++) {
        if( s[i] == '\n' || s[i] == '\0') {
            break;
        }
        if( s[i] == 'B' && s[i+1] ==  'W' ){
            count++;
            for( j = 1;  (i-j) >= 0; j++) {
                if( s[i-j] == 'B') {
                    count++;
                    s[i-j] = 'W';
                } else {
                    break;
                }
            }
            if( j == 1) {
                s[i] = 'W';
            }
            s[i+1] = 'B';
        }
    }
    printf("%lu",count);
}