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)

/*
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*/
#include <iostream>
#include <complex>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <numeric>
#include <cstring>
#include <ctime>
#include <cstdlib>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <cmath>
#include <bitset>
#include <cassert>
#include <queue>
#include <stack>
#include <deque>
       
        
using namespace std;
template<typename T1, typename T2>inline void chkmin(T1 &x, T2 y) { if (x > y) x = y; }
template<typename T1, typename T2>inline void chkmax(T1 &x, T2 y) { if (x < y) x = y; } 
template<typename T, typename U> inline ostream &operator<< (ostream &_out, const pair<T, U> &_p) { _out << _p.first << ' ' << _p.second; return _out; }
template<typename T, typename U> inline istream &operator>> (istream &_in, pair<T, U> &_p) { _in >> _p.first >> _p.second; return _in; }
template<typename T> inline ostream &operator<< (ostream &_out, const vector<T> &_v) { if (_v.empty()) { return _out; } _out << _v.front(); for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline istream &operator>> (istream &_in, vector<T> &_v) { for (auto &_i : _v) { _in >> _i; } return _in; }
template<typename T> inline ostream &operator<< (ostream &_out, const set<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline ostream &operator<< (ostream &_out, const multiset<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline ostream &operator<< (ostream &_out, const unordered_set<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline ostream &operator<< (ostream &_out, const unordered_multiset<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T, typename U> inline ostream &operator<< (ostream &_out, const map<T, U> &_m) { if (_m.empty()) { return _out; } _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')'; for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) { _out << ", (" << _it->first << ": " << _it->second << ')'; } return _out; }
template<typename T, typename U> inline ostream &operator<< (ostream &_out, const unordered_map<T, U> &_m) { if (_m.empty()) { return _out; } _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')'; for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) { _out << ", (" << _it->first << ": " << _it->second << ')'; } return _out; }
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define left left228
#define right right228
#define next next228
#define rank rank228
#define prev prev228
#define y1 y1228                                                         
#define read(FILENAME) freopen((FILENAME + ".in").c_str(), "r", stdin)
#define write(FILENAME) freopen((FILENAME + ".out").c_str(), "w", stdout)
#define files(FILENAME) read(FILENAME), write(FILENAME)
#define pb push_back
const string FILENAME = "input";



int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //read(FILENAME);
    string s;
    cin >> s;
    long long ans = 0;
    int curcnt = 0;
    for (int i = sz(s) - 1; i >= 0; i--) {
        if (s[i] == 'B') {
            ans += curcnt;
        } else {
            curcnt++;
        }
    }
    cout << ans << '\n';
    return 0;
}