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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <iostream>
#include <map>
#include <string>
#include <vector>
#include <time.h>     
using namespace std;
typedef long long int ll;
char S_c[1000000];
int main(int argc, char const *argv[]) {

  scanf("%s", S_c);
  size_t N = sizeof(S_c) / sizeof(S_c[0]);
  string S(S_c, N);
  vector<int> black_left{0};
  for (int i = 1; i < N; i++) {
    if (S[i - 1] == 'B') {
      black_left.push_back(black_left[i-1]+1);
    }else{
      black_left.push_back(black_left[i-1]);
    }
  }
  ll ans=0;
  for (int i = 1; i < N; i++) {
    if (S[i] == 'W') {
      ans+=black_left[i];
    }
  }
  cout << ans << endl;

  return 0;
}