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 <algorithm>
#include <iomanip>
#include <iostream>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define SORT(a) sort(a.begin(), a.end())
#define RSORT(a) sort(a.rbegin(), a.rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define RREP(i, n) for (int i = n; 0 <= i; i--)
#define FOR(i, start, end) for (int i = start; i < end; i++)
#define FORSTEP(i, start, end, step) for (int i = start; i < end; i += step)
#define RFOR(i, start, end) for (int i = start; end <= i; i--)
#define ALL(a) a.begin(), a.end()
#define MOD(a) a %= 1000000007
typedef long long ll;
using namespace std;
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    string s;
    cin >> s;

    int black = 0;
    REP(i, s.size()) if (s[i] == 'B') black++;
    int dist = s.size() - 1;
    vector<int> v(s.size());
    REP(i, s.size()) {
        if (s[i] == 'B') {
            v[i] = dist - i - black + 1;
            black--;
        }
    }

    ll ans = 0;
    REP(i, v.size()) { ans += v[i]; }

    std::cout << ans << std::endl;

    return 0;
}