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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program test
implicit none

integer(8) :: i,j,a(0:200001)=2,n,ans=0,b=0
character(len = 200000) :: s

read(*,*) s

n = len_trim(s)

do i= 1,n
	if(s(i:i) == "B")then
		a(i) = 0
	else
		a(i) = 1
	endif
enddo

do i= 1,n
	if(a(i) == 0) then
	b = b+1
	else
	ans = ans + b
	endif
enddo

write(*,*) ans










contains
subroutine heapsort_down(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) > y1(j+1)) j = j+1
				endif
				
				if (t1 > y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_down
 
subroutine heapsort_up(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) < y1(j+1)) j = j+1
				endif
				
				if (t1 < y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_up
 
end program