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)

import java.util.NoSuchElementException
import java.io.IOException

internal class FastScanner {
    private val `in` = System.`in`
    private val buffer = ByteArray(1024)
    private var ptr = 0
    private var buflen = 0
    private fun hasNextByte(): Boolean {
        if (ptr < buflen) {
            return true
        } else {
            ptr = 0
            try {
                buflen = `in`.read(buffer)
            } catch (e: IOException) {
                e.printStackTrace()
            }

            if (buflen <= 0) {
                return false
            }
        }
        return true
    }

    private fun readByte(): Int {
        return if (hasNextByte()) buffer[ptr++].toInt() else -1
    }

    private fun isPrintableChar(c: Int): Boolean {
        return c in 33..126
    }

    operator fun hasNext(): Boolean {
        while (hasNextByte() && !isPrintableChar(buffer[ptr].toInt())) ptr++
        return hasNextByte()
    }

    operator fun next(): String {
        if (!hasNext()) throw NoSuchElementException()
        val sb = StringBuilder()
        var b = readByte()
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b)
            b = readByte()
        }
        return sb.toString()
    }

    fun nextLong(): Long {
        if (!hasNext()) throw NoSuchElementException()
        var n: Long = 0
        var minus = false
        var b = readByte()
        if (b == '-'.toInt()) {
            minus = true
            b = readByte()
        }
        if (b < '0'.toInt() || '9'.toInt() < b) {
            throw NumberFormatException()
        }
        while (true) {
            if ('0'.toInt() <= b && b <= '9'.toInt()) {
                n *= 10
                n += (b - '0'.toInt()).toLong()
            } else return if (b == -1 || !isPrintableChar(b)) {
                if (minus) -n else n
            } else {
                throw NumberFormatException()
            }
            b = readByte()
        }
    }

    fun nextInt(): Int {
        val nl = nextLong()
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw NumberFormatException()
        return nl.toInt()
    }

    fun nextDouble(): Double {
        return java.lang.Double.parseDouble(next())
    }
}

fun main(args: Array<String>){
    val sc = FastScanner()
    val s = sc.next()
    var ans = 0
    var wCnt = 0
    for (i in 0 until s.length) {
        if (s[i] == 'W') {
            ans += i - wCnt
            wCnt++
        }
    }
    println(ans)
}