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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int main()
{

    char c[200000];
    int s[100001], sum = 0;
    scanf("%s", c);
    int len = strlen(c) - 1, k = 0, front = 0;

    for (int i = 0; i <= len; i++)
    {
        while (c[i] == 'B')
        {
            s[k]++;
            i++;
        }
        k++;
        while (c[i] == 'W')
        {
            s[k]++;
            i++;
        }
        i--;
        k++;
    }
    int end = k - 1;

    if (c[0] == 'W')
    {
        front++;
    }
    if (c[len] == 'B')
    {
        end--;
    }
    do
    {
        for (int i = front; i < end; i += 2)
        {
            sum += (s[i] + s[i + 1]) - 1;
        }
        front++;
        end--;
    } while ((front - end) != 1);

    printf("%d\n", sum);

    return 0;
}