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)

#include<bits/stdc++.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<iostream>
#include<vector>
#include<cstdlib>
#include<stdio.h>
#include<cmath>
#include<limits>
#include<iomanip>

using namespace std;
 
template <class T> using V = vector<T>;
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
const long long INF = 1LL << 60;
 
using ll = long long;
using db = long double;
using st = string;
using ch = char;
using bl = bool;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;
using vbl = V<bl>;
using graph = V<V<int>>;
using pq = priority_queue<ll>;


 
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define rFOR(i,a,b) for(ll i=(a);i>=(b);i--)
#define oFOR(i,a,b) for(ll i=(a);i<(b);i+=2)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define M(a,b) max(a,b)
#define rM(a,b) min(a,b)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define co(a) cout<<a<<endl;
#define pb(a) push_back(a);
  
 

ll sum(ll n) {
ll m=0;
  while(n){
    m+=n%10;
    n/=10;
  }
  return m;
}
 
 
const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
void Comuse() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
 
#define comuse Comuse()
 
ll combi(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
 
ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}


vll div(ll n){
  vll ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i==0){
      ret.push_back(i);
      if(i*i!=n){
        ret.push_back(n/i);
      }
    }
  }
  SORT(ret);
  return (ret);
}

void Solve();

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(20)<<fixed;
  Solve();
}


/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/
 

void Solve(){
  st s;
    cin>>s;
  ll n=s.sz;
  ll ans=0;
  //count white
  ll count=0;
  FOR(i,0,n){
    if(s[i]=='B'){
      count++;
    }
    else{
      ans+=count;
      //count=0;
    }
  }
  REV(s);
  count=0;
  FOR(i,0,n){
    if(s[i]=='W'){
      count++;
    }
    else{
      ans+=count;
    }
  }
  cout<<ans/2<<endl;
}