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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
  input! {
    n: usize,
    k: usize,
    s: String
  }
  let mut chars: Vec<char> = s.clone().chars().collect();
  
  let mut compresses: Vec<usize> = vec![];
  let mut consective = 1;
  let mut last = chars.remove(0);
  for c in chars {
    if c == last {
      consective += 1;
    } else {
      compresses.push(consective);
      last = c;
      consective = 1;
    }
  }
  compresses.push(consective);
  let len = compresses.len();
  if len <= k * 2 {
    println!("{}", s.len() - 1);
    return
  }
  
  let mut result = k * 2;
  for v in compresses {
    result += v - 1;
  }
  
  println!("{}", result);
}