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<algorithm>
#include<iostream>
#include<iterator>
#include<string>
#include<vector>

using namespace std;

int main()
{
    string S; cin >> S;
    vector<char> A;
    for(auto item : S) A.push_back(item);
    int a = 0;
    int k = 0;
    for(int i = 0; i < A.size(); i++)
        if(A[i] == 'B')
            a += i, k++;
    cout << static_cast<int>(0.5 * k * ((A.size()) + (A.size() - k -1))) - a << endl;
    return 0;
}