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 (Clang 3.8.0) (judge verdict: Accepted)

#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define SORT(a) sort(a.begin(), a.end())
#define ll long long
#define INF 1000000000000000000

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <sstream>
#include <cmath>
#include <ios>
#include <iomanip>

using namespace std;

//=================================================================
//Grand Contest 029 A
int main(){
    string S;
    cin >> S;
    ll b=0;
    ll w=0;
    ll ans=0;
    ll ans2=0;

    for(ll i=0; i< S.length(); i++){
        if(S[i]=='B'){
            b++;
        }else{
            ans += b;
            ans2 += i-w;
            w++;
        }
        //cout << ans << " " << ans2 << endl;
    }
    cout << ans2 << endl;
}