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 (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(clippy::many_single_char_names)]

// language: Rust(1.42.0)
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"
// My Library Repo' is at "https://github.com/Loptall/prcn_lib"

/*
    6月で水色になるぞ〜!!!
*/

// from std...
use std::cmp::{
    max, min,
    Ordering::{Equal, Greater, Less},
    Reverse,
};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::mem::swap;
use std::ops::*;
use std::process::exit;
use std::{f32, f64, i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

// Dep' crates are...
use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_complex::Complex;
use num_integer::{binomial, gcd, lcm, Integer};
use num_rational::Rational;
use num_traits::{clamp, one, pow, zero, One, Pow, Zero};
use permutohedron::Heap;
use petgraph::unionfind::UnionFind;
use proconio::{fastout, input, is_stdin_empty, marker::*};
use rand::random;

pub const MOD: usize = 1000000007;
pub const MAX: usize = 1usize << 63usize;
pub const MIN: i64 = i64::MIN;
pub const PI: f64 = f64::consts::PI;
pub const A: u8 = 'A' as u8;

/// ランレングス圧縮
pub fn run_length_encoding<T: PartialEq + Copy>(v: &Vec<T>) -> Vec<(T, usize)> {
    let mut res = Vec::new();
    let mut c = 1usize;
    let mut h = v[0];
    for e in v.iter().skip(1) {
        if h != *e {
            res.push((h, c));
            c = 1;
            h = *e;
        } else {
            c += 1;
        }
    }
    res.push((h, c));

    res
}

// code...
// #[fastout]
fn main() {
    input! {
        n: usize,
        mut k: usize,
        s: Chars
    }

    let rl = s.windows(2).filter(|x| x[0] == 'R' && x[1] == 'L').count().saturating_sub(1);
    let mut ans = run_length_encoding(&s)
        .into_iter()
        .map(|x| x.1 - 1)
        .sum::<usize>();

    dbg!(rl, ans);

    ans += 2 * min(k, rl);
    k -= min(k, rl);

    if k > 0 {
        ans += 1;
    }

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