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 (1.15.1) (judge verdict: Accepted)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}
fn adjacent4(x: usize, y: usize, sx: usize, sy: usize) -> Vec<(usize, usize)> {
    [(-1, 0), (1, 0), (0, -1), (0, 1)]
        .into_iter()
        .filter_map(|&(dx, dy)| {
            let nx = x as isize + dx;
            let ny = y as isize + dy;
            if nx >= 0 && nx < sx as isize && ny >= 0 && ny < sy as isize {
                Some((nx as usize, ny as usize))
            } else {
                None
            }
        })
        .collect()
}

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
fn main() {
    input! {
        s: chars,
    }
    // 最大の回数を出力する
    // bとwを交換してずらしていくと考えると, 最終的にwとbがそれぞれ並ぶことになる
    // wの個数分, bの個数分
    // bをずらす(wをずらす)回数は,（原点からのwiの距離) = (wiより手前にあるwの個数)
    // をwの個数分足すことでわかる
    // wの個数をカウントする変数とインデックスで計算すればいい

    let mut count = 0;
    let mut ans = 0;
    for i in 0..s.len() {
        if s[i as usize] == 'W' {
            ans += i - count;
            count += 1;
        }
    }
    println!("{}", ans);
}