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::cmp::max;
use std::collections::BinaryHeap;

fn read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let n: usize = read(&mut stdin_lock, &mut buf, b' ');
    let k: usize = read(&mut stdin_lock, &mut buf, b'\n');
    let s: String = read(&mut stdin_lock, &mut buf, b'\n');
    let cv: Vec<char> = s.chars().collect();

    let mut lc = 0i32;
    let mut rc = 0i32;
    for i in 0..n {
        if cv[i] == 'L' {
            lc += 1;
        } else {
            rc += 1;
        }
    }

    let mut lh = BinaryHeap::new();
    let mut rh = BinaryHeap::new();
    let mut c = 1;
    let mut d = cv[0];
    for i in 1..n {
        if cv[i] != d {
            if d == 'L' {
                lh.push(c);
            } else {
                rh.push(c);
            }
            c = 1;
            d = cv[i];
        } else {
            c += 1;
        }
    }
    if d == 'L' {
        lh.push(c);
    } else {
        rh.push(c);
    }

    let mut lmax = lc;
    for _ in 0..k {
        if rh.is_empty() {
            break;
        }
        lmax += rh.pop().unwrap();
    }
    lmax -= rh.len() as i32 + if cv[0] == 'L' { 1 } else { 0 } + if !rh.is_empty() && cv[n - 1] == 'R' { -1 } else { 0 };

    let mut rmax = rc;
    for _ in 0..k {
        if lh.is_empty() {
            break;
        }
        rmax += lh.pop().unwrap();
    }
    rmax -= lh.len() as i32 + if !lh.is_empty() && cv[0] == 'L' { -1 } else { 0 } + if cv[n - 1] == 'R' { 1 } else { 0 };

    println!("{}", max(lmax, rmax));
}