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>
//#define int long long

using namespace std;
using LL = long long;

#define FOR(i, a, n) for(int i = (int)(a); i < (int)(n); ++i)
#define REP(i, n) FOR(i, 0, n)

#define pb(a) push_back(a)
#define all(x) (x).begin(),(x).end()

const int INF = (int)1e9;
const LL INFL = (LL)1e19;
const int MOD = (int)1e9 + 7;

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    string s;
    cin >> s;

    int ans = 0;
    int cnt = 0;
    int n = s.size();
    REP(i, n) {
        if (s[i] == 'B') cnt++;
        else ans += cnt;
    }
    cout << ans << '\n';
}