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>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define INFTY 999999

//char型のポインタ(=配列と意味は同じ)をString型と定義
typedef char* String;

//文字列の長さを求める関数
int string_len(String str){
    int len=0;
    while(str[len]!=0){
        len++;
    }
    return len;
}

//文字列を標準入力から読み込み、それをString型のメモリを確保し直してくれる関数
String string_input(void){
    int i,len;
    char buf[200001];
    String str;
    scanf("%s",buf);//buf=入力した文字列が入るchar型の配列
    //ここまでは文字列の長さがわからなかったから、一旦1000個分のメモリを用意しそこに入力させる
    
    len=string_len(buf);//len=文字列の長さを表すint型変数
    NEW(str,len+1);//strはただのポインタだから、ここでNEWを使ってメモリを確保しなくちゃいけない
    for(i=0;i<len;i++){
        str[i]=buf[i];
    }
    str[len]=0;
    
    return str;
}



int main(void){
    int n;
    String S;
    
    S=string_input();
    n=string_len(S);
    
    //Wが前からなんばんめか
    int cw=0;
    
    int ans=0;
    int d;
    
    for(int i=0;i<n;i++){
        if(S[i]=='W'){
            cw++;
            d=(i+1)-cw;
            ans += d;
        }
    }
    
    
    printf("%d\n",ans);
    
    return 0;
    
}