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: Time Limit Exceeded)

#include<stdio.h>

int main(void) {
	char st[20000];
	char B, W;
	fgets(st, 20000, stdin);
	int i, j = 0, n = 0;
	while(*(st+1) != '\0') {
		n++;
	}
	for(i = n; i > 0; i--) {
		if(*(st+i-1) == B && *(st+i) == W) {
			*(st+i-1) = W;
			*(st+i) = B;
			j++;
		}
	}
	printf("%d", j);
	return 0;
}