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>
using namespace std;
#define SORT(x) sort(x.begin(),x.end())
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
#define mod 1000000007
typedef long long ll;
const ll LINF = 1001002003004005006ll;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
#define No cout<<"No"<<endl
#define Yes cout<<"Yes"<<endl
#define NO cout<<"NO"<<endl
#define YES cout<<"YES"<<endl;
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main(){
    string s;
    cin>>s;
    ll n=s.size();

    //全部WWW...BBBにできる
    //編集距離を求める
    //片方を右に寄せることだけを考えればもう一方は自然と他方による

    ll b=0,w=0;
    rep(i,n){
        if(s[i]=='B')   b++;
        else            w++;
    }
    ll m=0,ans=0;
    for(ll i=0;i<w;i++){
        while(s[m]=='B'){
            m++;
        }
        ans+=abs(m-i);
        m++;
    }
    cout<<ans<<endl;
    return 0;
}