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 (1.15.1) (judge verdict: Accepted)

fn input() -> String {
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).unwrap();
	s.trim().to_owned()
}

fn input_c() -> std::io::Cursor<String> {
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).unwrap();
	std::io::Cursor::new(s.trim().to_owned())
}

fn read_t<T: std::str::FromStr>(r: &mut std::io::Read) -> T {
	use std::io::Read;
	use std::iter::Iterator;
	r.bytes().map(|b| b.ok().unwrap() as char).skip_while(|c| c.is_whitespace() == true).take_while(|c| c.is_whitespace() == false).collect::<String>().parse().ok().unwrap()
}

fn main() {
	let mut c = input_c();
    let     n = read_t::<usize>(&mut c);
    let     k = read_t::<usize>(&mut c);
    let     s = input();
    let     b = s.as_bytes();
    let mut d = 0;
    
    for i in 1..n {
    	if b[i - 1] != b[i] {
        	d += 1;
        }
    }
    
    if k * 2 <= d {
    	println!("{}", n - 1 - (d - k * 2));
    } else {
    	println!("{}", n - 1);
    }
}