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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> ii;
typedef map<int, int> MPII;
typedef set<int> SETI;

const int mxN = 2e5;
const ld pi = 4.0 * atanl(1.0);
const int iinf = 1e9 + 10;
const int inf = 1e18 + iinf + 10;
const int mod = 1000000007;
const ld prec = .000001;

#define F first
#define S second
#define pb push_back
#define mp make_pair
#define all(c) c.begin(), c.end()
#define rall(c) c.end(), c.begin()

int n, m, t, a[mxN];
vi v;

void fast() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); }

int main() {
  fast();
  string s;
  cin >> s;
  char curr = s[0];
  int ans = 1;
  for (int i = 1; i < s.length(); ++i) {
    if (s[i] != curr) ++ans, curr = s[i];
  }
  cout << ans;
  return 0;
}