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)

use std::io::Write;

#[allow(unused)]
macro_rules! debug {
    ($($format:tt)*) => (write!(std::io::stderr(), $($format)*).unwrap());
}
#[allow(unused)]
macro_rules! debugln {
    ($($format:tt)*) => (writeln!(std::io::stderr(), $($format)*).unwrap());
}

fn main() {
    let (n, k): (usize, usize) = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        let s = s.trim_right().to_owned();
        let mut ws = s.split_whitespace();
        let a = ws.next().unwrap().parse().unwrap();
        let b = ws.next().unwrap().parse().unwrap();
        (a, b)
    };

    let s: String = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        let s = s.trim_right().to_owned();
        s
    };

    let chars: Vec<_> = s.chars().collect();

    let mut prev = 'a';
    let mut happiness = 0;
    let mut groups = 0;
    for &c in chars.iter() {
        if prev == c {
            happiness += 1;
        } else {
            prev = c;
            groups += 1;
        }
    }
    debugln!("{}", happiness);
    debugln!("{}", groups);

    let tmp = std::cmp::min(k, groups / 2);
    if groups % 2 == 0 && tmp == groups / 2 {
        println!("{}", happiness + tmp * 2 - 1);
    } else {
        println!("{}", happiness + tmp * 2);
    }
    // if tmp % 2 == 0 && tmp == k {
    //     debugln!("!!if");
    //     println!("{}", happiness + ((tmp - 1) / 2) * 2 + 1);
    // } else {
    //     debugln!("!!else");
    //     println!("{}", happiness + (tmp / 2) * 2);
    // }
}