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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int MOD = 1e9 + 7;
typedef vector<ll> vll;
typedef vector<int> vi;
const int maxn = 100 + 10;

int main() {
    string s;
    cin >> s;
    int ans = 0;
    int left = 0;
    int right = 0;
    for (int i = 0; i < s.length(); ++i) {
        if (s[i] == 'B') {
            left = i;
            break;
        }
    }
    for (int i = left + 1; i < s.length(); ++i) {
        if (s[i] == 'W') {
            right = i;
            break;
        }
    }

    while (left < s.length() && right < s.length()) {
        while (right < s.length() && s[right] == 'B') ++right;
        if (right >= s.length()) break;
        while (left < right && s[left] == 'W') ++left;
        if (left >= right) {
            ++right;
            continue;
        }
        ans += right - left;
        swap(s[left], s[right]);
        ++left;
        ++right;
    }
    cout << ans << endl;
    return 0;
}