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: Time Limit Exceeded)

#include <stdio.h>
int length,SUM;
int choose;
int digui(char S[],int whiltNumber){
	int i ,choose = 0;;
	for (i =length - 1; i >=0; i-- ){
		if (S[i] == 'W' && S[i-1] == 'B'){
			S[i] = 'B';
			S[i-1] = 'W';
			SUM ++;
		}
	}
	for (i = 0; i < length; i++){
		if (S[i] == 'W'){
				choose ++;
		} else {
			if (choose == whiltNumber) {
				return 1;
			}
			break;
		}
	}
	return 0;
}

int main () {
	char S[200000];
	int i,whiltNumber=0;
	gets(S);
	for(i = 0; ;i++){
		if (S[i] == '\0') {
			length = i;
			break;
		}
		if (S[i] == 'W') {
			whiltNumber++;
		}
	}

	do
	{
		if (digui(S,whiltNumber) == 1) {
			break;
		}
	} while (1);

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

	return 0;
}