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: Runtime Error)

#include<stdio.h>
#define N 100000
int main()
{
    int i,count=0,n;
    char s[N];
    char *p=s;
    gets(s);
    int y;
    for(;*p!='\0';p++)n++;
    do{
    y=-1;
    for(i=0;i<n;i++)
    {
        if(s[i]=='B'&&s[i+1]=='W')
        {
            s[i+1]='B';
            s[i]='W';
            count++;
            y=1;
        }
    }
    }while(y==1);
    printf("%d",count);
    return 0;
}