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)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    
    let s = get().as_bytes();
    
    let mut ans = 0;
    
    // 交換を眺めてみたら転倒数ぽい
    // 転倒数の求め方ググってもよく分からん
    
    let mut c = 0_i64;
    for &ch in s.iter() {
    	if ch == b'B'  {
        	c += 1;
        } else if ch == b'W' {
        	ans += c;
        }
    }
    println!("{}", ans);
}