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: Wrong Answer)

fun a1(args: Array<String>) {
    var bcnt = 0
    var ans = 0
    args[0].forEach {
        if (it == 'B') {
            bcnt++
        } else {
            ans += bcnt
        }
    }
    println(ans)
}

fun main(args: Array<String>) {
    a1(readLineList())
}

fun readLineList(): Array<String> {
    val lineList = mutableListOf<String>()
    while (true) {
        val line = readLine()
        if (line.isNullOrBlank()) {
            break
        } else {
            lineList.add(line!!)
        }
    }
    return lineList.toTypedArray()
}