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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i < (n); ++i)
using namespace std;
typedef long long ll;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    string s;
    cin >> s;
    int n = s.size();

    int end = n;
    for(int i = n-1; i >= 0; --i){
        if(s[i] == 'B') --end;
        if(s[i] == 'W') break;
    }
    n = end;

    int res = 0;
    rep(i, n){
        if(s[i] == 'B') {
            res += end - i - 1;
            --end;
        }
    }

    cout << res << endl;

    return 0;
}