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>

#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define mp make_pair 
#define F first
#define S second
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define Yes(hoge) cout<<((hoge)?"Yes":"No")<<endl;
#define YES(hoge) cout<<((hoge)?"YES":"NO")<<endl;
typedef struct aaaaa{
  int x,y,t;
}Grid;
using namespace std;

int main(){
  string s;cin>>s;
  vector<ll>hog(s.size());
  for(int i=0;i<s.size();i++){
    if(s[i]=='B')
      hog[i]=1;
    else
      hog[i]=0;
  }
  for(int i=1;i<s.size();i++){
    hog[i]+=hog[i-1];
  }
  ll ans=0;
  for(int i=0;i<s.size();i++){
    if(s[i]=='W')
      ans+=hog[i];
  }
  cout<<ans<<endl;
  return 0;
}