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: Wrong Answer)

use std::str::FromStr;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn calculate_larger_group(s: &Vec<char>) -> (usize, usize){

    let mut i = 0;
    let mut num_of_r = 0;
    let mut num_of_l = 0;
    while i < s.len(){
        let mut prev_i = i;
        while i < s.len() && s[i] == 'L'{
            i += 1;
        }
        if prev_i != i{
            num_of_l += 1;
        }
        prev_i = i;
        while  i < s.len() && s[i] == 'R'{
            i += 1;
        }
        if prev_i != i{
            num_of_r += 1;
        }        
    }
    return (num_of_l, num_of_r);
}

fn get_sum_hapiness(s: &Vec<char>) -> usize{
    let (lg, rg) = calculate_larger_group(s);
    if lg == 0 || rg == 0{
        return s.len() - 1;
    }else{
        return s.len() - 1 - (lg + rg-1);
    }

}

fn get_larger_char(s: &Vec<char>) -> char{
    let s = calculate_larger_group(s);

    if s.0 >= s.1{
        return 'L';
    }else{
        return 'R';
    }
}

fn main(){
    let line: Vec<usize> = read_as_vec();
    let (n, mut k) = (line[0], line[1]);

    let mut s = read_as_string().chars().collect::<Vec<char>>();
    //println!("{:?}", s);
    let larger_c: char = get_larger_char(&s);

    let mut i = 0;
    //replace
    while i < s.len() && k > 0{
        if s[i] != larger_c{
            k -= 1;
        }
        while i < s.len() && s[i] != larger_c{
            s[i] = larger_c;
            i += 1;
        }

        //just skip
        while i < s.len() && s[i] == larger_c{
            i += 1;
        }
    }
    println!("{}", get_sum_hapiness(&s));
    //println!("{:?}", s);
}