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 <cstring> 
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
int main()
{
	int num1=0,num2=0,sum=0,sum3=0;
	string s;
	cin>>s;
	bool pd=true;
	for(int i=0;i<s.length();i++)
	{
		if(s[i]=='B')
		{
			if(pd)
			{
				num1++;
				pd=false;
				cout<<"&";
			} 	
		}
		/*if(s[i]=='W'&&pd)
		{
			sum3++; 
			cout<<sum3<<"&";
		}*/
		if(s[i]=='W'&&pd==false)
		{
			pd=true;
			num2=i+1;
			sum=sum+num2-num1;
		}
		
	}
	cout<<sum;
	return 0;
}