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: Time Limit Exceeded)

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()

    fun reverse(array: Array<Boolean>): Int {
        val new = Array(array.size) { array[it] }
        var done = false
        var start = 0
        for (i in 0 until array.size - 1) {
            if (done) break

            if (array[i] && !array[i + 1]) {
                done = true
                start = if (i < 2) 2 else i

                new[i] = !array[i]
                new[i + 1] = !array[i + 1]
            }
        }

        return if (done) reverse(new.sliceArray(start - 2 until array.size)) + 1 else 0
    }

    println(reverse(S))
}