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: Time Limit Exceeded)

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <math.h>
int main()
{   int i,n,cnt=0;
    char s[200002];
    scanf("%s",s);
    int a[200002];
    n=strlen(s);
    for(i=0;i<n;i++){
        if(s[i]=='B'){
            a[i]=0;
        }else{
            a[i]=1;
        }
        
    }int chk=1;
    while(chk==1){
        chk=0;
    for(i=0;i<n-1;i++){
        if(a[i]==0&&a[i+1]==1){
            a[i]=1;a[i+1]=0;
            chk=1;
            cnt++;
        }
    }
    }
    printf("%d",cnt);

    return 0;
}