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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main()
{
    string S;
    cin>>S;
    int N=S.size();
    int B[N];
    long long count=0;
    B[0]=0;
if(N>=1&&N<=2*100000){
    if(S[0]=='B'){
        B[0]+=1;
    }
    else{
        B[0]=0;
    }
    for(int i=1;i<N;i++){
        B[i]=B[i-1];
        if(S[i]=='B')
            B[i]+=1;
    }
    for(int i=0;i<N;i++){
        if(S[i]=='W'){
        count+=(i-1>=0?B[i-1]:0);
    }
    }
    cout<<count;
}
    return 0;
}