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 <cstdio>
#include <vector>
#include <algorithm>
#include <limits.h>
#include <queue>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
#define REP(i, n) for(int i=0; i<n; i++)

bool black[300100];
int main(){
    string S;
    cin >> S;
    int N = S.length();
    REP(i,N){
        char x = S[i];
        if(x == 'B')
            black[i] = true;
        else
            black[i] = false;
    }

    int num = 0;
    int ans = 0;
    for(int i = N-1; i >= 0; i--){
       int T = N-1-num;
       if(black[i]){
           num++;
           ans += T - i;
       }
    }
    cout << ans << endl;
    return 0;
}