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: Time Limit Exceeded)

#include <iostream>
#include <vector>
using namespace std;

int main() {
	int i = 0;
	vector<bool> v;
	char c;
	while(cin >> c){
		v.push_back(c=='B');
	}
	int count = -1;
	int countsum = 0;
	while(count != 0){
	count = 0;
	for(int i = 0;i<v.size()-1;i++){
		if(v[i]){
			if(!v[i+1]){
				v[i] = false;
				v[i+1] = true;
				count++;
			}
		}
	}
	countsum += count;
	}
	cout << countsum << endl;
	return 0;
}