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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
	char S[20002];

	scanf("%s", S);

	int N = strlen(S);

	// 置換回数
	int count = 0;

	// 直前のBの連続回数
	int Bs = 0;

	int i = 0;
	while(i < N){
		if(S[i] == 'B'){
			Bs++;
		}
		else{ // 'W'
			count += Bs;
			
			S[i-Bs] = 'W';
			S[i] = 'B';
			
//			printf("Index = %d\n", i);
//			printf("Bs    = %d\n", Bs);
//			printf("Count = %d\n", count);
//			printf("%s\n", S);
			
//			Bs = Bs;
		}

		i++;
	}

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

	return 0;
}