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: Wrong Answer)

/*
    ##############################
    #  Author: Pratyush Gaurav   #
    #  College: NIT ROURKELA     #
    ##############################
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long lli;
typedef long double ld;
typedef pair<lli, lli > plli;
typedef vector<lli> vlli;
typedef vector<plli > vplli;

#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a * b) / gcd(a, b))
#define sqr(x) (x)*(x)
#define all(a) a.begin(),a.end()
#define UN(v) sort(all(v)), v.resize(unique(all(v))-v.begin())

#define endl '\n'

const long long INF = 100000000000000000;
const long long MOD = 1000000007;
const long long MAXN = 100005;

lli dx[]={0,0,-1,1,-1,-1,1,1};
lli dy[]={1,-1,0,0,1,-1,-1,1};

void solve() {
    string s;
    cin >> s;
    int pos = 0, ans = 0;
    vlli v;
    for (int i = 0; i < s.length(); ++i) {
        if (s[i] == 'W') {
            v.push_back(i);
        }
    }
    for (auto i : v) {
        ans += (i - pos);
        pos++;
    }
    cout << ans << endl;
}

int main()
{
    IOS;
    lli t = 1;
    //cin >> t;

    while(t--)
        solve();
    return 0;
}