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<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<list>
#include<numeric>

using namespace std;

int main(void){

  string s;

  cin>>s;

  //cout << s.length() << endl;
  char c;
  int val=0;
  vector<int> result(s.length());
  int last_B=s.length();
  vector<int> res_b(0);
  int tmp=0;
  for(int i=0;i<s.length();i++){
    if(s[i] == 'B'){
      tmp++;
    }
    else if(s[i] == 'W'){
      res_b.push_back(tmp);
    }
  }


  cout << (int)accumulate(res_b.begin(),res_b.end(),0) << endl;

  return 0;
}