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 main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();
    let k: usize = iter.next().unwrap().parse().unwrap();

    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let s = buf.trim().chars().collect::<Vec<char>>();

    let mut s_n = Vec::new();
    let mut prev_c = s[0];
    let mut cnt_c = 1;
    for &c in s[1..].iter() {
        if c != prev_c {
            s_n.push(cnt_c);
            prev_c = c;
            cnt_c = 1;
        } else {
            cnt_c += 1;
        }
    }
    s_n.push(cnt_c);

    let nr_part = 2 * k + 1;

    let mut ans = 0;

    if nr_part >= s_n.len() {
        ans = n - 1;
    } else {
        /*
        let mut sum = s_n[..nr_part].iter().fold(0, |s, &x| s+x);
        ans = sum;
        for i in nr_part..s_n.len() {
            sum = sum + s_n[i] - s_n[i-nr_part];
            ans = std::cmp::max(ans, sum);
        }
        */
        let sum = s_n.iter().fold(0, |s, &x| s+x-1);
        ans = sum + 2 * k;
    }

    println!("{}", ans);
}