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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<stdio.h>
#include<set>
using namespace std;
int main(){


    string s;
    cin>>s;
    int n=s.size();
    if(s.size()==0){
        cout<<0<<endl;
        return 0;
    }

    int os[n+1];
    for(int i=0;i<n+1;i++)os[i]=0;
    int j=1;
    int  ans=0;
    for(int i=n;i>=0;i--){
        if(s[i]=='B'){
            os[j]=os[j-1];
            //cout<<"b"<<os[j]<<endl;
            ans+=os[j];
            j++;
        }else if(s[i]=='W'){
            os[j]+=os[j-1]+1;
            //cout<<"w"<<os[j]<<endl;
            j++;
        }
    }
    
    cout<<ans<<endl;

return 0;
}