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: Wrong Answer)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<bitset>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#include<iomanip>
using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define INF 2e9
#define MOD 1000000007
#define LINF (long long)4e18

using ll = long long;
using Pi = pair<int,int>;
using Pl = pair<ll,ll>;



int main(){
   string S; cin >> S;
   vector<int> sum(S.size()+1);
   for(int i = S.size(); i >= 1; i--){
       if(S[i-1] == 'W') sum[i-1] = sum[i] + 1;
       else sum[i-1] = sum[i];
   }
   int ans = 0;
   rep(i,S.size()){
       if(S[i] == 'B') ans += sum[i+1];
   }
   cout << ans << endl;

}