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 <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main(){
    string S;
    cin >> S;

    long long black_pos = 0;
    for(size_t i = 0; i < S.size(); i++){
        if(S[i] == 'B') black_pos += i;
    }
    sort(S.rbegin(), S.rend());
    long long new_black_pos = 0;
    for(size_t i = 0; i < S.size(); i++){
        if(S[i] == 'B') new_black_pos += i;
    }
    cout << new_black_pos - black_pos << endl;
    /*
    long long white_num = 0;
    long long count = 0;
    for(int i = 0; i < (int)S.size(); i++){
        if(S[i] == 'W'){
            count += i - white_num;
            white_num++;
        }
    }
    cout << count << endl;
    */
}