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: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <iomanip>
#include <sstream>

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#define _USE_MATH_DEFINES
#define elif else if
typedef long long ll;
using namespace std;
const int INF = 1e8;
//const double EPS = 1e-15;
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	//freopen ("auxiliary.in", "w", stdin);
	//freopen ("auxiliary.out", "r", stdout);

	//ifstream fin;
	//fin.open(".in");
	//ofstream fout(".out");

	string s;
	cin >> s;
	ll ans = 0, cur = 0;
	for (int i = s.size() - 1; i >= 0; --i) {
		if (s[i] == 'W')
			++cur;
		else
			ans += cur;
	}
	cout << ans;
	return 0;
}