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: Wrong Answer)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

fn main() {
    input! {
        mut s:Chars,
    }
    let mut ans = 0;
    for _ in 0..50 {
        let mut prevb = false;
        for i in 0..s.len() {
            if prevb && s[i] == 'W' {
                s[i] = 'B';
                ans += 1;
                prevb = true;
            }
            if i + 1 < s.len() && s[i] == 'B' && s[i + 1] == 'W' {
                s[i + 1] = 'B';
                s[i] = 'W';
                ans += 1;
                prevb = false;
            }
            if prevb && s[i] == 'W' {
                s[i] = 'B';
                ans += 1;
                prevb = true;
            }
            if i + 1 < s.len() && s[i] == 'B' && s[i + 1] == 'W' {
                s[i + 1] = 'B';
                s[i] = 'W';
                ans += 1;
                prevb = false;
            }
            prevb = s[i] == 'B';
        }
    }
    println!("{}", ans);
}