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)

use std::cmp::*;
use std::io::*;
fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let s: Vec<char> = itr.next().unwrap().chars().collect();
    let n = s.len();
    let mut t = vec![0u64; n + 1];
    for i in 0..n {
        if s[i] == 'B' {
            t[i] += 1;
        }
        t[i + 1] += t[i];
    }
    let mut ans = 0u64;
    for i in 0..n {
        if s[i] == 'W' {
            ans += t[i];
        }
    }
    println!("{}", ans);
}