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>
#define all(vec) vec.begin(),vec.end()
using namespace std;
using ll=int_fast64_t;
using P=pair<ll,ll>;
const ll INF=1LL<<30;
const ll LINF=1LL<<61;
const double EPS=1e-9;
const int MOD=1e9+7;
int main(){
    string s;cin>>s;
    int n=s.length();
    ll c=0;
    ll ans=0;
    for(int i=0;i<n;i++){
        if(s[i]=='B'){
            c++;
        }else{
            ans+=c;
        }
    }
    cout<<ans<<endl;
}