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 text[200001];
	text[0] = '\0';
	int len=0;
	int Bcount=0;
	int Wcount=0;
	int a=0;
	int b=0;
	long int answer=0;
	scanf("%s",text);
	len = strlen(text);
	for (int i = 0; i < len; i++){
		if (strncmp(&text[i],"B",1) == 0){
			Bcount++;
		}
	}
	Wcount = len - Bcount;
	//printf("%d %d %d\n",len,Bcount,Wcount);
	if (Bcount >= Wcount){
		a = Wcount;
	for (int i = 0; i < len; i++){
		if (strncmp(&text[i],"B",1) == 0){
			answer += (a - i);
			//printf("%d\n",answer);
			a++;
		}
	}
}
	if (Bcount < Wcount){
	for (int i = 0; i < len; i++){
		if (strncmp(&text[i],"W",1) == 0){
			answer += i - b;
			//printf("%d %d\n",answer,i);
			b++;
		}
	}
}
	//printf("%s\n",text);
	printf("%d\n",answer);
	return 0;
}