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 <bits/stdc++.h>
using ll = long long;
using namespace std;
ll GCD(ll a, ll b){ return a ? GCD(b % a, a) : b; }
int main(){
    string s;
    cin>>s;
    ll ans=0;
    vector<int> bw;
    bool b=s[0]=='B';
    for(int i=0; i<s.size(); i++){
        int cnt=0;
        for(int j=i; j<s.size(); j++){
            if(s[j]=='B' && b) cnt++;
            else if(s[j]=='W' && !b) cnt++;
            else{
                i=j-1;
                break;
            }
        }
        bw.push_back(cnt);
        b=!b;
    }
    int i=0;
    int bcnt=0;
    if(s[0]=='W') i++;
    for(i; i<bw.size()-1; i+=2){
        bcnt+=bw[i];
        ans+=bcnt*bw[i+1];
    }

    cout<<ans<<endl;
}