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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define fin(ans) cout << (ans) << endl;
#define blank(ans) cout << (ans) << " ";
#define mp(p, q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()

typedef long long ll;
typedef long long lli;
typedef unsigned long long ull;
typedef long double ld;
typedef string str;

typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<ld> vd;
typedef vector<str> vs;
typedef vector<ll> vll;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<ll, str>> vpls;
typedef vector<tuple<str, ll, ll>> vtsl;
typedef vector<vector<ll>> vvll;
typedef vector<vector<char>> vvc;
typedef vector<vector<str>> vvs;

const ld PI = acos(-1.0);
const ll MAX = 9000000000000000000;
const ll MIN = -9000000000000000000;
const ld DMAX = 4500;
const ld DMIN = -4500;
const ll MOD = 1000000007;

//実行部
void Main()
{
  str s; cin >> s;
  ll flag = 0;
  while(1)
  {
    ll q = 0;
    for(ll i = 1;i < s.size();i++)
    {
      if(s[i - 1] == 'B'&&s[i] == 'W')
      {
        s[i - 1] = 'W';
        s[i] = 'B';
        flag++;
        q++;
      }
    }
    if(q == 0)
    {
      fin(flag)
      return;
    }
  }
  return;
}

//前処理
int main()
{
  //ios::sync_with_stdio(false);
  //cin.tie(0);
  cout << fixed << setprecision(20); //高精度少数表示
  Main();
  return 0;
}