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 (Clang 3.8.0) (judge verdict: Wrong Answer)

 #include <stdio.h>
 
int o=0;
int main(void){
    
    char a[200005];
    int sum=0,b[200005];
    
    for(int i=0;i<=200005;i++){
        scanf("%s\n",&a[i]);
    }
    
    
    for(int i=0;i<=200005;i++){
        if(a[i] == 'W'){
            b[o]=i;
            o++;
        }
    }
    for(int i=0;i<=o;i++){
            sum += b[i]-i;
        }
    printf("%d\n",sum+o);
}