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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.awt.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[]$) {
        Scanner sc = new Scanner(in);
        String s=sc.next();
        int c=0;
        int ans=0;
        int stop=s.length();
        for (int i = 0; i <stop ; i++) {
            if(s.charAt(i)=='W'){
                ans+=c;
            }else{
                c++;
            }
        }
        out.println(ans);
    }
    static long power(long x,int n){
        long mod=1000000007;
        long ans=1;
        while (n>0) {
            if((n&1)==1){
                ans=(ans*x)%mod;
            }
            x=(x*x)%mod;
            n>>=1;
        }
        return ans;
    }
    //最大公約数
    static int gcd (int a,int b) {
        int temp;
        while((temp = a%b)!=0) {
            a = b;
            b = temp;
        }
        return b;
    }
    //※最小公倍数はa*b/gcd(a,b)である
    static class UF{
        static int size=51;
        static int[] par=new int[size];
        //初期化
        static void init(){
            for (int i = 1; i <size ; i++) {
                par[i]=i;
            }
        }
        //根をたどる
        static int root(int x){
            if(par[x]==x){
                return x;
            }else{
                return par[x]=root(par[x]);
            }
        }
        //同じ根をもつか(同じ集合に属するか)判定
        static boolean same(int x,int y){
            return root(x)==root(y);
        }
        //合体
        static void unite(int x,int y){
            x=root(x);
            y=root(y);
            if(x==y)return;
            par[x]=y;
        }
    }
}