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 <bits/stdc++.h>

#ifdef NON_SUBMIT
#define TEST(n) (n)
#else
#define TEST(n) ((void)0)
#endif

using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	TEST(freopen("input.txt", "r", stdin));
	TEST(freopen("output.txt", "w", stdout));
	TEST(freopen("debug.txt", "w", stderr));
	string s;
	int R;
	long long ans = 0;
	cin >> s;
	R = s.size() - (s.back() == 'B');
	for (int i = s.size() - 2; i >= 0; i--) {
		if (s[i] == 'B') {
			if (s[i + 1] == 'W') {
				ans += --R - i;
				s[i] = 'W';
			}
			else {
				R = i;
			}
		}
	}
	cout << ans << '\n';
	return 0;
}