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>
int main(){
 int c=0,b=0,w=0,n,wc=0;
 int be,af;
 char s[200000];
 scanf("%s",s);
 while(s[c]!=0){
  c++;
  if(s[c]=='b')b=b+c;
  if(s[c]=='w')w=w+c,wc++;
 }
 n=c;
 be=abs(b-w);
 b=0,w=0;
 for(c=0;c<n;c++){
  if(c<wc)w=c+w;
  else b=b+c;
 }
 af=abs(w-b);
 if(af-be>0)printf("%d\n",af-be);
 else printf("0\n");
 return 0;
}