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 <iostream>
#include <string>
unsigned int ans(0);
bool isEnd = false;
unsigned int Wnum = 0;

int main() {
	std::string S;
	std::getline(std::cin,S);
	S.push_back('\0');
	while (!isEnd) {
		bool checking = false;
		unsigned int addnum = 0;
		unsigned int ignoreWnum = Wnum;
		for (int i = 0;i < S.length();i++) {
			if (S[i] == '\0') {
				isEnd = true;
			}
			if (checking) {
				if (S[i] == 'B') {
					addnum++;
				}
				else {
					Wnum ++;
					break;
				}
			}
			else {
				if (S[i] == 'B') {
					if (ignoreWnum == 0) {
						checking = true;
					}
					addnum++;
					ignoreWnum--;
				}
			}
		}
		if (!isEnd) {
			ans += addnum;
		}
	}
	std::cout << ans << '\n';
	return 0;
}