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>
#include <numeric>
using namespace std;
#define ll long long
#define mod107 1000000007
#define rep(i, n) for (int i = 0; i < n; ++i)
using Graph = vector<vector<int>>;
#define grepq priority_queue<double, std::vector<double>, std::greater<double>>
#define all(v) v.begin(), v.end()
#define PI acos(-1)
 
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}
const ll INF = 10010001010;
const int inf = 1000034000;
 
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
 
int main() {
  string S;
  cin >> S;
  int N = S.size();
  int black = 0;
  int ans = 0;
  for(int i = 0;i < N;i++){
    if(S[i] == 'B'){
      black++;
    }
    else{
      ans += black;
    }
  }
  cout << ans << endl;
  
}