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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#define moji 1000000 

char s[moji];
int i=0;

void cmoji(int l);

int main(){
    int l;
    int v=1;
    scanf("%s",&s);
    while(v==1){
        v=0;
        for(l=1;l<moji;l++){
            if(s[l]=='W'&&s[l-1]=='B') {
                 i++;
                 cmoji(l);
                 v=1;
            } 
        }
    }
    printf("%d",i);
    return 0;
}

void cmoji(int l){
    s[l]='B';
    s[l-1]='W';
}