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<algorithm>
#include<vector>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstdlib>
#include<utility>

typedef long long ll;
#define INF 10e9

using namespace std;

ll dp[200005];

int main() {
	string s;
	cin >> s;

	ll buf = 0;
	for (int i = 0; s[i] != 'B'; i++) {
		buf++;
	}
	cout << buf << endl;
	for (int i = buf + 1; i < s.size(); i++) {
	    if (s[i] == 'W' && s[i - 1] == 'B') {
			dp[i] = dp[i - 1] + i - buf;
			s[i] = 'B';
			buf++;
		}
		else
			dp[i] = dp[i - 1];
		for (int j = 0; j < s.size(); j++) cout << dp[j] << " ";
		cout << endl;
	}
	cout << dp[s.size() - 1] << endl;
}