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: Runtime Error)

// A - Irreversible operation
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define uli unsigned long long int
#define rep(i, m, n) for (lli i = m; i < (n); i++)
#define repe(i, m, n) for (lli i = m; i <= (n); i++)
#define ALL(x) (x).begin(), (x).end()
#define SIZE(x) ((lli)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define _GLIBCXX_DEBUG
const lli INF = 2147483647;
const lli MINF = -2147483648;
const lli LINF = 9223372036854775807;
const lli MOD = 1000000007; //10^9+7
const double PI = acos(-1);
#define SORT(n) sort(n.begin(), n.end())
#define SORTR(n) sort(n.begin(), n.end(), greater<int>())
#define REV(n) reverse(n.begin(), n.end())
#define pb push_back
#define eb emplace_back
#define mp make_pair
#pragma GCC optimize("Ofast")
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vlli = vector<lli>;
using vs = vector<string>;
using ll = long long;
// vector入力
template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
  for (T &x : vec)
    is >> x;
  return is;
}
// vector出力
template <typename T>
ostream &operator<<(ostream &os, vector<T> &vec)
{
  // os << '{';
  for (int i = 0; i < vec.size(); i++)
  {
    os << vec[i] << (i + 1 == vec.size() ? "" : "");
  }
  // os << '}';
  return os;
}
// aよりもbが大きいならばaをbで更新する(更新されたならばtrueを返す)
template <typename T>
bool chmax(T &a, const T &b)
{
  if (a < b)
  {
    a = b; // aをbで更新
    return true;
  }
  return false;
}
// aよりもbが小さいならばaをbで更新する(更新されたならばtrueを返す)
template <typename T>
bool chmin(T &a, const T &b)
{
  if (a > b)
  {
    a = b; // aをbで更新
    return true;
  }
  return false;
}

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  lli a, b, c, h, k, n, w, ans = 0, count = 0;
  string s, t, v, x, y;
  vector<pair<string, pair<lli, lli>>> pr;
  cin >> s;
  vlli d(110, 0), e(s.size() + 1, 0);
  rep(j, 0, s.size())
  {
    y = s[j];
    if (y == "W")
    {
      d[j]++;
    }
    e[j + 1] = e[j] + d[j];
  }
  a = MAX(e);
  rep(i, 1, s.size())
  {
    x = s[i - 1];
    if (x == "B")
    {
      ans += a - e[i];
    }
  }
  // cout << d << endl;
  // cout << e << endl;
  cout << ans << endl;
  // cout << "s: " << s << endl;
}