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 <bits/stdc++.h>
using namespace std;

int main() {
  string s; cin >> s;
  
  vector<int> stones(s.size(), 0);
  for (int i = 0; i < s.size(); i++) if (s.at(i) == 'B') stones.at(i) = 1;
  
  vector<int> ps_stones(s.size(), 0);
  partial_sum(stones.begin(), stones.end(), ps_stones.begin());
  
  int answer = 0;
  for (int i = 0; i < s.size(); i++) if (s.at(i) == 'W') answer += ps_stones.at(i);
  
  cout << answer << endl;
}