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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
#define rg register
#define ll long long
#define LDB long double
#define ull unsigned long long
#define view(i,x) for(rg int i=hd[x];i!=-1;i=e[i].nt)
#define go(i,x,a) for(rg int i=a;i<x;i++)
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
using namespace std;

const int maxn=2*1e5+5;
ll ans;
char s[maxn];

inline int rd(){
	int ret=0,af=1; char gc=getchar();
	while(gc < '0' || gc > '9'){ if(gc=='-') af=-af; gc=getchar(); }
	while(gc >= '0' && gc <= '9') ret=ret*10+gc-'0',gc=getchar();
	return ret*af;
}

int main(){
	scanf("%s",s);
	int n=strlen(s); int tt=0;
	go(i,n,0){
		if(s[i] == 'W') ans+=tt;
		else tt++;
	}
	printf("%lld",ans);
	return 0;
}//Faze