Problem:
Score : 400 points

Problem Statement

There are N people standing in a queue from west to east.

Given is a string S of length N representing the directions of the people.
The i-th person from the west is facing west if the i-th character of S is L, and east if that character of S is R.

A person is happy if the person in front of him/her is facing the same direction.
If no person is standing in front of a person, however, he/she is not happy.

You can perform the following operation any number of times between 0 and K (inclusive):

Operation: Choose integers l and r such that 1 \leq l \leq r \leq N, and rotate by 180 degrees the part of the queue: the l-th, (l+1)-th, ..., r-th persons. That is, for each i = 0, 1, ..., r-l, the (l + i)-th person from the west will stand the (r - i)-th from the west after the operation, facing east if he/she is facing west now, and vice versa.

What is the maximum possible number of happy people you can have?

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

|S| = N

Each character of S is L or R.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of happy people after at most K operations.

Sample Input 1

6 1
LRLRRL

Sample Output 1

3

If we choose (l, r) = (2, 5), we have LLLRLL, where the 2-nd, 3-rd, and 6-th persons from the west are happy.

Sample Input 2

13 3
LRRLRLRRLRLLR

Sample Output 2

9

Sample Input 3

10 1
LLLLLRRRRR

Sample Output 3

9

Sample Input 4

9 2
RRRLRLRLL

Sample Output 4

7

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using u64 = unsigned long long;
#define REP(i, n) for (int i = 0; (i64)(i) < (i64)(n); ++i)

#ifdef ENABLE_DEBUG
template <typename T>
void debug(T value) {
  cerr << value;
}
template <typename T, typename... Ts>
void debug(T value, Ts... args) {
  cerr << value << ", ";
  debug(args...);
}
#define DEBUG(...)                              \
  do {                                          \
    cerr << " \033[33m (L" << __LINE__ << ") "; \
    cerr << #__VA_ARGS__ << ":\033[0m ";        \
    debug(__VA_ARGS__);                         \
    cerr << endl;                               \
  } while (0)
#else
#define debug(...)
#define DEBUG(...)
#endif

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  i64 N, K;
  cin >> N >> K;
  string S;
  cin >> S;

  i64 unhappy = 0;
  for (int i = 0; i < N - 1; ++i) {
    if (S[i] == 'R' && S[i + 1] == 'L') {
      if (K > 0) {
        --K;
        for (int j = i + 1; j < N; ++j) {
          if (S[j] == 'R') {
            i = j - 1;
            break;
          }
          S[j] = 'R';
        }
      } else {
        unhappy += 2;
      }
    }
  }
  if (S[N - 1] == 'R') ++unhappy;
  if (S[0] == 'L' && K == 0) ++unhappy;
  cout << (N - unhappy) << endl;
}