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 <algorithm>
#include <stdio.h>
#include <deque>
using namespace std;
using ll = long long;

class Int2 {
public:
	ll color, count;
	Int2(ll color, ll count):color(color),count(count){}
};
int main() {
	char s[200001];
	scanf("%s", s);

	deque<Int2> c;
	c.emplace_back(s[0], 1);
	for (int i = 1; s[i]; i++) {
		if (c.back().color == s[i])
			c.back().count++;
		else
			c.emplace_back(s[i], 1);
	}
	if (c.back().color == 'B')
		c.pop_back();
	if (c.front().color == 'W')
		c.pop_front();

	ll count = 0, carry = 0;
	for (int i = 0; i + 1 < c.size(); i += 2) {
		c[i].count += carry;
		count += c[i].count * c[i + 1].count;
		carry = c[i].count;
	}
	printf("%lld\n", count);
}