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 Kotlin (1.0.0) (judge verdict: Accepted)

private val scanner = java.util.Scanner(System.`in`)
 
private fun readInts(n: Int) = Array(n) { scanner.nextInt() }
private fun readLongs(n: Int) = Array(n) { scanner.nextLong() }
private fun readLineInts() = readLine()!!.split(" ").map { it.toInt() }.toTypedArray()
private fun readLineLongs() = readLine()!!.split(" ").map { it.toLong() }.toTypedArray()
 
fun main(args: Array<String>) {
    val S = readLine()!!.map { it == 'B' }.toTypedArray()
    val size = S.size
 
    val sum = Array(size) { 0L }
    for (i in 0 until size) sum[i] += (if (S[i]) 1 else 0) + (if (i == 0) 0 else sum[i - 1])
 
    var count = 0L
    for (i in 0 until S.size) if (!S[i]) count += sum[i]
 
    println(count)
}