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 Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::{cmp::Reverse, collections::*};

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Chars, Usize1},
};

#[allow(unused_imports)]
use itertools::Itertools;

#[allow(unused_imports)]
use num::Integer;

#[fastout]
pub fn main() {
    input! {
        s: Chars,
    }
    let b = s.iter().position(|c| *c == 'B');
    if let Some(b) = b {
        let tmp: Vec<_> = s
            .iter()
            .enumerate()
            .skip(b)
            .filter(|(_, c)| **c == 'W')
            .map(|(i, _)| i - b)
            .collect();
        println!(
            "{}",
            tmp.iter().enumerate().map(|(i, j)| j - i).sum::<usize>()
        );
    } else {
        println!("0");
    }
}