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>

char in[(2*10*10*10*10*10*10) + 1] = {0};

int main(void)
{
	int counter = 0;
	int b_count = 0;
	int i = 0;

	scanf("%s", &in[0]);
	for(i = 0; i < (2*10*10*10*10*10*10) ; i++) {
		if (in[i] == 'B') {
			b_count++;
			continue;
		}
		if (in[i] == 'W') {
			counter += b_count;
		}
		if(in[i+1]==0){
			break;
		}
	}
    printf("%d\n",counter);
    return 0;
}