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)

#include <iostream>
#include <cmath>
#include <map>
using namespace std;

int show_list(map<long, int> &m)
{
    cout << "list: ";
    for(auto itr = m.begin(); itr != m.end(); itr++) {
        for(int i = 0; i < itr->second; i++) {
            cout << itr->first << " ";
        }
    }
    cout << endl;
}

int main()
{
    int N, tmp;
    int d = 0;
    map<long, int> m;
    int ans = 0;

    cin >> N;
    for(int i = 0; i < N; i++) {
        cin >> tmp;
        m[tmp]++;
    }

    for(auto itr = m.rbegin(); itr != m.rend(); itr++) {
        //show_list(m);
        long v = itr->first;
        //cout << "v: " << v << endl;
        for(int t = 1; t <= 32; t++) {
            long val = pow(2, t);
            //cout << "val = " << val << endl;
            if(v <= val && val < v * 2) {
                while(m.find(v) != m.end() && m[v] > 0 && m.find(val - v) != m.end() && m[val - v] > 0) {
                    //cout << "hit = " << v << ", " << val - v << endl;
                    m[val - v]--;
                    m[v]--;
                    ans++;
                }
            } else if(val == v * 2) {
                while(m.find(v * 2) != m.end() && m[v * 2] > 1) {
                    //cout << "hit = " << v << ", " << v << endl;
                    m[v] -= 2;
                    ans++;
                }
            }
        }
    }

    cout << ans << endl;

    return 0;
}