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)

#![allow(unused_imports, unused_macros, dead_code)]

macro_rules! min {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| min!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = min!($($ys),*);
        if $x < t { $x } else { t }
    }}
}
macro_rules! max {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| max!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = max!($($ys),*);
        if $x > t { $x } else { t }
    }}
}


macro_rules! ewriteln {
    ($($args:expr),*) => { let _ = writeln!(&mut std::io::stderr(), $($args),*); };
}
macro_rules! trace {
    ($x:expr) => { ewriteln!(">>> {} = {:?}", stringify!($x), $x) };
    ($($xs:expr),*) => { trace!(($($xs),*)) }
}
macro_rules! put {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        if let Some(x) = it.next() { print!("{}", x); }
        for x in it { print!(" {}", x); }
        println!("");
    }};
    ($x:expr) => { println!("{}", $x) };
    ($x:expr, $($xs:expr),*) => { print!("{} ", $x); put!($($xs),*) }
}

const M: i64 = 1_000_000_007;

fn main() {
    let mut sc = Scanner::new();
    let n: i64 = sc.cin();
    let k: i64 = sc.cin();
    let s: Vec<char> = sc.cin::<String>().chars().collect();

    let mut g: i64 = 0;
    {
        let mut last = 'a';
        for &c in s.iter() {
            if last == c { continue }
            last = c;
            g += 1;
        }
    }
    g = max!(g - 2 * k, 1);
    put!(n - g);

}

use std::io::{self, Write};
use std::str::FromStr;
use std::collections::VecDeque;

struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
impl Scanner {
    fn new() -> Self { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn cin<T: FromStr>(&mut self) -> T {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
        self.buffer.pop_front().unwrap().parse::<T>().ok().unwrap()
    }
}