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 <utility>
#include <vector>
#include <iostream>
#include <cstdlib>
#include <bits/stdc++.h>
#include <limits>
#include <set>
#include <map>
#include <string>
#include <sstream>
#include <stdio.h>
#define REP(i,n) for(int i=0;i<n;i++)
typedef long long ll;
typedef unsigned long long ull;
ull gcd(ull a, ull b){return b? gcd(b, a%b):a;}
using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
	string S;
	cin >> S;

	int blackCount = 0;
	int ans = 0;
	for(int i = 0; i < S.length(); i++){
		if(S[i] == 'B'){
			blackCount++;
		}else if(S[i] == 'W'){
			ans += blackCount;
		}
	}
	cout << ans << endl;

    return 0;
}